ELEMENTS 

OP 


MACHINE    DESIGN 


BY 

HENRY  L.  NACHMAN 

Associate  Professor  of  Kinematics  and  Machine  Design, 
Armour  Institute  of  Technology,  Chicago,  Illinois 


FIRST  EDITION 


NEW  YORK 

JOHN  WILEY  &  SONS,  INC. 

LONDON:  CHAPMAN  &  HALL,  LIMITED 

1918 


Copyright,  1918 

BY 

HENRY  L.  NACHMAN 


PRESS  OF 

BRAUNWORTH   A   CO. 

BOOK  MANUFACTURERS 

BROOKLYN,   N.   V. 


TABLE  OF   CONTENTS 


CHAPTER  . PAGE 

I.  STRENGTH  OF_MATERIALS 1 

FASTENINGS 

II.  SCREW  FASTENINGS 19 

III.  RIVETED  JOINTS 32 

IV.  KEYS  AND  COTTERS 46 

V.  SHRINK  AND  FORCE  FITS 64 

TRANSMISSION  MACHINE  PARTS 

VI.  SHAFTS  AND  AXLES 57 

VII.  COUPLINGS  AND  CLUTCHES 64 

VIII.  JOURNALS  AND  BEARINGS 75 

IX.  BELTS  AND  PULLEYS 93 

X.  FRICTION  WHEELS 107 

XI.  TOOTHED  GEARS Ill 

XII.  ROPE  TRANSMISSION 121 

XIII.  CHAIN  GEARING 134 

XIV.  PIPES  AND  CYLINDERS 137 

XV.  VALVES 148 

XVI.  FLY-WHEELS 155 

XVII.  CRANK-SHAFTS,  CRANK-PINS,  AND  ECCENTRICS 164 

XVIII.  CONNECTING  RODS,  PISTON  RODS,  AND  ECCENTRIC  RODS 179 

XIX.  PISTONS,  CROSS-HEADS  AND  STUFFING-BOXES 191 

XX.  HOISTING  MACHINERY  DETAILS 203 

XXI.  SPRINGS 216 

XXII.  MATERIALS  OF  MACHINERY 224 

iii 


423688 


PREFACE 


THIS  little  volume  is  intended  primarily  as  a  class-room  text 
book  on  the  subject  of  elementary  machine  design.  It  is  the 
result  of  the  author's  experience,  extending  over  more  than  fifteen 
years,  both  in  practical  design  and  in  teaching  of  this  subject. 
The  pre-requisites  on  the  part  of  the  student  are  thorough  courses 
in  machine  drawing  and  elementary  mechanics. 

Based  on  the  brief  outline  of  the  strength  of  materials  given 
in  Chapter  I  the  author  has  attempted  to  develop  the  equa- 
tions for  the  design  of  the  more  common  machine  elements. 
This  has  generally  been  done  very  concisely  and  frequently  only 
an  outline  of  the  deduction  has  been  given.  Empirical  formulae 
and  rule  of  thumb  methods,  so  much  used  in  elementary  texts 
on  this  subject,  have  been  avoided  as  far  as  possible.  There  are 
many  factors  which  in  practice  affect  the  design  of  machine  parts 
that  cannot  be  discussed  profitably  in  the  class-room ;  for  instance, 
cost  of  construction,  capacity  of  shop  machinery,  etc.  For  this 
reason  the  teacher  must  be  content  if  the  student  acquires  the 
power  to  analyze  the  forces  and  the  resultant  stresses  in  machine 
parts  and  to  apply  the  proper  equations  for  their  design. 

The  illustrations  have  been  carefully  chosen  to  show  typical 
constructions  rather  than  a  great  variety  which  tend  to  confuse 
the  inexperienced  student.  The  standard  text  books  as  well  as 
the  American  and  European  technical  press  have  been  freely 
consulted  in  the  preparation  of  the  manuscript. 

H.  L.  NACHMAN. 
CHICAGO,  March,  1918, 


ELEMENTS  OF  MACHINE  DESIGN 


ERRATA 

Page  11.     Last  figure  upside  down. 

12.     Line  14,  for  pin  ends  read  flat  ends. 
12.     Line  15,  for  flat  ends  read  pin  ends. 

59.     Equation  (7)  should  read  A3/— . 

\  ss 

62.     Prob.  1,  for  tension  read  torsion. 
92.     Prob.  3,  for  500,000  read  50,000. 

116.     Fig.   11-4,   pitch  radius  Ri  at  large  end  of  gear  is 
omitted. 

V2 
160.     Equation  (2)  should  read  c*  =  T7r  (nearly). 

169.     Equation  (22)  should  read  T^R^h+li) -P14. 
171.     Equation  (28)  should  read  b-     =  L 


K/f 

177.  Equation  (34)  should  read  s  =  —  = 

z       2 

178.  In  sketch  for  prob.  4  crank  radius  =  18". 
204.     Equation  (1)  should  read  <  =  .02Z)+f". 
211.     Equation  (9)  for  Pb  read  Pt. 

211.     Equation  (10)  for  T  read  Pt. 


it  is  gradually  increased  there  will  be  reached  a  point  where  if  the 
load  is  removed  the  bar  does  not  resume  its  original  form.  This 
ooint  is  called  the  elastic  limit  of  the  material.  If  the  load  is  still 
further  increased  rupture  of  the  material  will  occur  sooner  or 


ELEMENTS  OF  MACHINE  DESIGN 


CHAPTER  I 
STRENGTH  OF  MATERIALS 

Elasticity.  When  an  external  force  acts  on  a  (rigid)  body 
it  produces  a  change  of  shape  in  that  body.  This  deformation 
is  called  strain.  When  the  force  is  removed  the  body  returns 
to  its  original  shape.  This  property  which  enables  a  body  to 
resume  its  initial  shape  after  the  application  of  an  external  force 
is  called  the  elasticity  of  the  material. 

Load.  The  external  forces  acting  on  a  machine  part  or 
structure  are  called  loads.  A  dead  load  being  one  which  has  a 
constant  value  while  a  live  load  is  one  which  continually  changes 
in  value.  Thus  the  load  on  a  column  supporting  the  roof  of  a 
building  is  a  dead  load,  while  the  forces  acting  on  the  connecting 
rod  of  a  steam  engine  constitute  a  live  load.  If  the  load  be  sud- 
denly applied  or  removed  the  body  carrying  such  a  load  is  said 
to  be  subjected  to  shock. 

Stress.  When  a  body  is  strained  by  the  application  of  an 
external  force  there  are  set  up  internal  forces  between  the  particles 
or  molecules  which  constitute  it,  which  hold  the  external  forces  in 
equilibrium.  These  internal  forces  are  called  stresses.  The  nature 
and  magnitude  of  the  stresses  depends  on  the  method  of  appli- 
cation of  the  external  forces. 

Factor  of  Safety.  If  a  load  be  applied  to  a  bar  of  metal  and 
it  is  gradually  increased  there  will  be  reached  a  point  where  if  the 
load  is  removed  the  bar  does  not  resume  its  original  form.  This 
ooint  is  called  the  elastic  limit  of  the  material.  If  the  load  is  still 
further  increased  rupture  of  the  material  will  occur  sooner  or 


FVMAGHINE  DESIGN 


later.  The  load  at  which  such  failure  takes  place  is  called  the 
ultimate  load  and  the  stress  produced  the  ultimate  stress.  In 
practice  of  course  no  part  of  a  machine  should  be  loaded  to  any- 
where near  its  ultimate  load.  The  actual  load  which  the  part  may 
be  permitted  to  carry  is  called  the  safe  load  and  the  stress  pro- 
duced by  this  load  the  safe  stress.  The  ratio  of  ultimate  to  safe 
stress  is  called  the  factor  of  safety  (F)  or 


„_  ultimate  stress 
safe  stress 


In  the  choice  of  a  proper  factor  of  safety  the  machine  designer 
must  use  a  good  deal  of  judgment,  as  its  value  varies  between 
very  wide  limits,  depending  on  the  material,  the  kind  of  load 
and  the  purpose  of  machine.  The  effect  of  live  loads  and  shocks 
is  to  weaken  the  material,  and  experience  has  shown  that  the  ratio 
of  the  factor  of  safety  of  any  material  under  dead  load,  live  load 
and  shock  should  be  approximately  1:2:3;  thus,  if  the  factor  of 
safety  for  any  material  under  dead  load  is  3,  then  6  would  be  a 
correct  value  for  a  live  load  and  9  for  shock. 

Modulus  of  Elasticity.  If  a  bar  be  subjected  to  a  constantly 
increasing  load  it  will  be  found  that  within  the  elastic  limit  the 
amount  of  strain  is  proportional  to  the  load.  That  is,  if  a  load 
of  1000  Ibs.  be  applied  to  the  bar  and  it  is  stretched  .001  in.  a 
load  of  2000  Ibs.  will  stretch  it  .002  in.  The  ratio  of  the  stress 
produced  by  a  load  to  the  strain  is  called  the  modulus  of  elasticity 
and  is  a  constant  for  any  one  material,  viz.  : 

Modulus  of  elasticity  =  —    — , (1) 

strain 

this  ratio  usually  being  denoted  by  the  letter  E. 

Tension.    There  are  three  kinds  of  simple  stresses, 
viz.:  tension,  compression  and  shear.     If  a  bar  be 
»p          loaded  by  a  force  P,  acting  parallel  to  the  axis  of 
FIG.  1-1.       the  bar  (Fig.  1-1)  it  produces  an  elongation,  tensile 
strain,    of   the   bar.      If  A   be  the  cross-sectional 
area  of  the  bar  and  L  its  length  then,  since  the  unit  of  strain  is 


STRENGTH  OF  MATERIALS  3 

always  designated  per  unit  (inch)  of  length,  and  stress  per  unit 
(square  inch)  of  area,  we  have 

total  elongation 
strain = j—^ , 

stress  =  -r. 
A 

If  we  denote  the  actual  or  safe  stresses  for  any  material  by  s 
and  ultimate  stresses — those  at  which  failure  of  material  occurs — 
by  U  or  more  particularly  safe  and  ultimate  tensile  stresses  by 
st  and  Ut)  we  have 


or 

P  =  Ast,     .     .    ......     (3) 

or  since  F  =  —  -  this  may  also  be  written 

(3a) 


In  this  equation  P  will  be  the  safe  load.     If  the  breaking  load 
be  required  we  have 

P=AUt  .........     (4) 

If  strain  be  denoted  by  5  we  have  the  equation  for  modulus 
of  elasticity. 

stress     s      P  . 

=      ....... 


EXAMPLE  1.  A  wrought  iron  tube  2  in.  external  and  1J  in. 
internal  diameter  supports  a  tensile  load  of  12,000  Ibs.  Find 
stress  in  pounds  per  square  inch  and  total  elongation  of  tube  if 
E  =  29,000,000  and  tube  is  8  ft.  long,  also  factor  of  safety 

Cross-sectional  area  of  tube  is 


=  1.38  sq.ins. 


4  ELEMENTS  OF  MACHINE  DESIGN 

Therefore  stress  is 


P     12000    o-™,, 

!=T38~  per  sq'm* 


Factor  of  safety  is 


From  (5)  we  have 

12000 


29000000X1.38 
=  .0003  in. 

This  is  the  elongation  per  inch  of  length  and  therefore 
total  elongation  =  8  X  12  X  .0003 
=  .0288  in. 


Compression.  If  a  block  be  loaded  by  a  force  P 
parallel  to  its  axis  in  such  a  way  as  to  shorten  it 
(Fig.  1-2);  P  constitutes  a  compressive  load  and 
produces  compressive  stresses  and  strains  in  the 
material.  Using  the  same  notation  with  the  addi- 


FIG  1-2        ^on  °^  ^c  ^Or  ^e  ultimate  compressive  stress  and 
sc  for  safe  or  actual  compressive  stress  we  have 


(7) 

and  since 


STRENGTH  OF  MATERIALS  5 

EXAMPLE  2.  The  estimated  weight  of  a  building  is  840  tons. 
This  weight  is  to  rest  on  twenty  brick  piers.  If  a  stress  of  150  Ibs. 
per  square  inch  be  allowed,  what  is  the  size  of  these  piers,  assum- 
ing them  to  have  a  square  section? 

Load  on  each  pier  is 


=  84000  Ibs. 

Cross-sectional  area  of  pier  from  (6)  is 

P    84000 


8e 


=  560  sq.ins. 


Therefore  sides  are  equal  t< 

V560  =  23.6  ins.,  say  24X24  ins. 

Shear.  If  two  equal  and  opposite  forces  act  in  the  same  plane 
tending  to  slide  the  sections  of  the  body  on  each  side  of  the 
plane  in  opposite  directions  they  produce  shearing  stresses  and 


FIG.  1-3. 

strains  in  that  plane.  Thus  in  Fig.  1-3  the  forces  P  produce 
shearing  stresses  and  strains  in  the  section  AB  of  the  rivet.  De- 
noting by  ss  and  Us  the  safe  and  ultimate  shearing  stresses, 


P 

Sa  =  -r. 

(9) 

A 
P=Ass  

MO) 

P-4& 

.  (11} 

6 


ELEMENTS  OF  MACHINE  DESIGN 


TABLE  1 


Material. 

MODULUS  OF  ELASTICITY. 

ULTIMATE  STRESS. 

Tension  or 
Compression. 

E 

Shear. 
Es 

Tension. 

Ut 

Compres- 
sion. 
Uc 

Shear. 

Us 

Cast  iron 

a4,ooo,ooo 

26,000,000 
30,000,000 
32,000,000 

6,000,000 
10,000,000 
12,000,000 

18,000 
50,000 
70,000 
120,000 
120,000 
24,000 
32,000 
20,000 
58,000 
60,000 
20,000 

90,000 
38,000 
80,000 

65,000 
12,000 

120,000 
12,000 

6800 
7000 
5400 
6300 
5700 

25,000 
40,000 
50,000 

24,000 

43,000 
12,000 

Wrought  iron  
Steel,  mild. 

Steel  tool 

Steel  wire 

Copper,  cast  
Copper,  hard  drawn.  . 
Brass 

12,000,000 
15,000,000 
9,000,000 
14,000,000 



Phosphor-bronze  .... 
M  anganese-bronze  . 

Aluminum,  cast  
Aluminum,  rolled.  .  .  . 

Ash. 

9,000,000 
10,000,000 

Average 
for 
timber 
1,500,000 

Average 
across 
grain 
400,000 

Oak  

Yellow  pine.      .    .  .  < 

Red  pine 

Spruce               .... 

Compound  Stresses.  Besides  these  simple  stresses  there  are 
three  compound  stresses,  viz.:  bending,  buckling,  and  twisting 
stresses.  If  a  bar  be  supported  at  one  or  both  ends,  Fig.  1-4, 

and  a  load  P  is  applied  at  right 
angles  to  its  axis,  it  will  produce 

I  )         bending  stresses.     If  a  bar  which 

is  long  in  comparison  to  its  other 
dimensions  be  supported  at  one 
or  both   ends    and   loaded  by  a 
FIG.  1-4.  force   P  parallel  to  its  axis,  it 

will    produce    buckling    stresses. 

(Fig.  1-5.)  A  bar  held  at  one  point  and  acted  upon  by  a  couple 
which  tends  to  turn  it  upon  its  axis  is  subjected  to  twisting  as 
shown  in  Fig.  1-6. 

Bending.  If  a  beam  of  rectangular  section  be  subjected  to 
bending  as  in  Fig.  1-7  it  will  take  the  curved  form  shown.  The 
material  in  top  of  beam  will  be  compressed,  that  in  bottom. 


STRENGTH  OF  MATERIALS 


1 


stretched,  while  that  in  center  will  be  neither  stretched  nor 
compressed.  This  plane  along  which  there  is  no  strain  and  there- 
fore no  stress  is  called  the  neutral 
plane.  This  plane  passes  through 
the  gravity  axes  of  the  sections 
of  beam. 

The  pressures  exerted  by  the 
supports  upon  the  beam  are 
called  the  reactions.  They  can 
in  the  simpler  cases  of  loading 
be  readily  determined  by  equat- 
ing the  sum  of  the  moments  of 
all  external  forces  acting,  about 
one  of  the  supports,  to  zero. 
Thus  in  Fig.  1-8,  let  I  be  the  dis- 
tance between  points  of  support, 
called  the  span  of  beam,  PI  and 
P<2.  are  the  loads,  RA  and  RB  are  the  reactions  at  the  supports; 


FIG.  1-5. 


FIG.  1-6. 


f 


AL 


JB 


FIG.  1-7. 


FIG.  1-8. 


then  taking  moments  about  A  and  equating  their  sum  to  zero 
we  obtain: 


and 


RB  = 


The  other  reaction  may  be  found  by  taking  moments  about  B 
or  more  simply  from  the  fact  that  the  sum  of  the  reaction  equals 
the  sum  of  the  loads,  thus 


8 


ELEMENTS  OF  MACHINE  DESIGN 


Bending  Moment.  Imagine  any  section  (Fig.  1-9)  distant 
x  from  one  of  the  supports.  If  moments  of  all  the  external  forces 
on  one  side  of  this  section  be  taken  about  the  section  the  sum  of 
these  moments  is  called  the  bending  moment  at  that  section  and 
will  be  designated  by  M .  Thus  in  Fig.  1-9  the  external  forces 
to  the  left  of  section  at  x  are  RA,  PI  ancl  P^  and  the  bending 
moment  is 

M  =  RAx-Pi(x-li)-P2(x-l2). 

The  bending  moments  will  thus  be  different  for  various  sections 
along  the  beam.  The  position  at  which  M  is  a  maximum  is 
called  the  dangerous  section  because  this  is  the  point  at  which 
the  stress  will  be  a  maximum  if  the  beam  be  one  having  a  uniform 


Is- 


FIG.  1-9. 


FIG.  1-10. 


section   throughout   its   length.     With    concentrated   loads   this 
is  always  under  one  of  the  loads. 

The  maximum  stress  which  occurs  in  any  one  section  is 
given  by  the  equation 

-^.  (12) 


In  this  equation  .7  is  the  moment  of  inertia  of  the  section  about 
its  neutral  axis  and  e  is  the  distance  from  the  neutral  axis  to 
outside  of  section.  If  Fig.  1-10  represent  the  section  of  a  rect- 
angular beam  and  the  small  area  a  be  multiplied  by  its  dis- 
tance ?/,  from  the  neutral  axis  xx,  squared,  the  sum  of  all  the  small 
areas  which  constitute  the  section  multiplied  by  the  square  of 
their  respective  distances  from  the  axis  xx  is  called  the  moment 
of  inertia,  /,  of  the  section.  The  moment  of  inertia,  I,  divided 
by  the  distance  e,  is  called  the  section  modulus  and  will  be  denoted 


STRENGTH  OP  MATERIALS 


9 


by  the  letter  z.     Table  2  gives  moments  of  inertia  of  various 
sections  together  with  their  section  moduli.     From  equation  (12), 


IT 
M=-, 


sz. 


(13) 


The  use  of  this  equation  is  most  easily  shown  by  an  example. 

EXAMPLE  3.     A  beam  of  10  ft.  span  carries  a  load  of  4000  Ibs. 

4  ft.  from  left  support  (Fig.  1-11).     The  beam  has  a  square  sec- 


FIG.  1-11., 


tion.    Find  dimensions  of  sections  so  that  stress  will  not  exceed 
1500  Ibs.  per  square  inch. 


Reaction  at  A 


2400  Ibs. 


Dangerous  section  is  under  load  and  bending  moment  at  this 
section  is 

Af  =  2400X4X12  =  115200  in  Ibs. 

M  =  sz. 
From  table  2   z  =  63 


X 115200 
1500 


=  7.7  ins.  say  8  ins. 


A  load  acting  at  one  point  on  a  beam  is  called  a  concentrated 
load,  while  if  it  is  spread  out  over  the  entire  beam  or  portions  of 
it,  it  is  called  a  distributed  load.  Concentrated  loads  will  be 
designated  by  P,  and  distributed  loads  by  Q.  Table  3  gives 
data  necessary  for  the  solution  of  all  beam  problems  which  are 
likeiy  to  be  met  with  in  the  subject  of  machine  design. 


10 


ELEMENTS  OF  MACHINE  DESIGN 
TABLE  2 


Section. 


Moment  of  Inertia. 


bH* 
12 


Section  Modulus. 
z 


bH* 
6 


12 


K— D— »H 


64 


BH3-bh* 
12 


BH*-bh* 
QH 


z=-  or 


12 


BW-bh* 
QH 


1 


12 


QH 


STRENGTH  OF   MATERIALS 
TABLE  3 


11 


Beam. 


Max.  Bending 
Moment. 


PL 


PL 
4 


Dangerous 
Section. 


At  support 


At  middle 


Deflection. 


PIS 

3EI 


PL3 


PL 

8 


At  A  and  B 


PL* 
192EI 


QL 
2 


At  support 


SEI 


QL 

8 


At  middle 


384^7 


PL 

24 


At.  A  and 


384^7 


12  [CLEMENTS  OF  MACHINE  DESIGN 

Buckling.  The  commonest  example  of  pieces  subjected  to 
buckling  stresses  are  columns.  The  strength  of  a  column  depends 
largely  on  how  its  ends  are  held.  In  some  cases  it  is  not  possible 
to  predetermine  whether  a  piece  will  fail  by  buckling  or  by  com- 
pression. It  is  then  of  course  necessary  to  determine  the  safe  load 
for  both  and  use  the  smaller  value.  Table  4  gives  safe  loads  for 
various  types  of  columns.  These  equations  are  known  as  Euler's 
column  equations. 

Euler's  equations  are  suitable  only  for  very  long  columns, 
that  is  such  in  which  the  length  is  not  less  than  about  45  times  the 
smallest  dimension  of  the  cross-section.  As  the  majority  of 
struts  and  columns  used  in  machine  construction  are  shorter 
another  equation  will  be  found^more  suitable;  this  is  known  as 
the  Rankine  equation.  For  flat-ends  it  is 


P  —         "C-fJ- 

/ 
A 

for  pin  ends 


A  =  area  of  section  of  column, 
k  —  a  constant  determined  by  experiment, 
/  =  the  least  moment  of  inertia  of  section. 

Substituting  the  proper  values  of  /  we  have  for  column  of 
circular  section  of  diameter  d. 

(Flat  ends)  n       ScA 


(Pin  ends) 


STRENGTH  OF  MATERIALS 

TABLE  4 


13 


Column. 


m 


Held  at  one 
end  free  at  the 
other. 


P  i  n  ends 
guided  along 
original  axis. 


Held  at  one 
end,  the  other 
end  guided. 


Both   ends 
held. 


Safe  Load  P. 


P  = 


Fl* 


Fl* 


NOTE. — /  =  least  mof 
ment  of  inertia  o- 
column  section. 


14  ELEMENTS  OF  MACHINE  DESIGN 

'For  rectangular  sections,  the  smallest  dimensioning  denoted 
by  b,  this  becomes 

(Round  ends)  n    '   ScA 


(Fla   ends) 


The  value  of  k  for 
Mild  steel  is 
Hard  steel  is 
Wrought  iron  is 
Cast  iron  is 
Timber  is 


EXAMPLE.  What  is  the  safe  load  on  a  strut  having  a  rect- 
angular section  2  by  4  ins.  and  a  length  of  5  ft?  The  material 
is  mild  steel  and  the  safe  stress  is  8000  Ibs.  per  square  inch. 
Round  ends  are  to  be  assumed. 

I     60 

As  ^  =  -^-  =  30  the  Rankine  equation  is  to  be  used.     Then 
o      A 

p  =    scA         8000  X  (2X4) 

!+*£    i+_J__M_2 
jp       T7500    (2)2 

12  12 

=  26200  Ibs. 

Torsion.  The  principal  stress  induced  by  torsion  is  a  shearing 
stress  and  will  therefore  be  denoted  by  ss.  The  force  which 
turns  or  tends  to  turn  the  shaft,  multiplied  by  its  perpendicular 
distance  to  center  of  shaft,  Fig.  1-12,  is  called  the  twisting  moment 
or  torque  and  will  be  denoted  by  T.  Thus 

T  =  PR. 


STRENGTH  OF   MATERIALS 


15 


The  stress  induced  in  a  rod  by  a  torque  T  is  given  by  the 

equation 

'f~ 

(14) 


and  therefore 


_Te 

ss  —  T  ) 
IP 


rp  _^j_P_ 

e 


=  sszp (15) 

In  this  equation  Ip  is  the  polar  moment  of  inertia  and  zp  is 
the  polar  section  modulus.  Table  5  gives  values  of  Ip  and  zp 
for  the  usual  sections. 


FIG.  1-12. 


The  angle  5  through  which  a  rod  subjected  to  torsion  is  turned 


is  given  by  the  equation 


=  57.2 


"T.L. 
IPES 

TL 


(16) 


Equations  (14),  (15),  and  (16)  are  true  strictly  only  for  cir- 
cular sections  but  may  be  used  with  sufficient  accuracy  for  other 
sections. 

Combined  Stresses.  A  machine  part  is  frequently  subjected 
to  both  bending  and  torsion.  In  this  case  it  is  usual  to  figure 
an  ideal  twisting  moment  which  if  applied  would  give  a  stress 
equivalent  to  the  actual  stress  due  to  bending  and  torsion.  This 
ideal  twisting  moment  (Tt}  is  then  used  in  equation  (15)  to  find 
the  stress.  The  value  of  it  is  calculated  from  the  equation 


16 


ELEMENTS  OF  MACHINE  DESIGN 
TABLE  5 


Section. 


Polar  Moment 
of  Inertia. 


Polar  Section 
Modulus. 

z, 


N D 


16       D 


STRENGTH  OF   MATERIALS 


17 


There  are  other  cases  of  combined  stresses  but  they  will  be  dis- 
cussed whenever  they  occur  in  the  text. 


PROBLEMS 

1.  How  much  will  a  steel  tube  1^  in.  outside  diameter  and  1  in.  inside 
diameter  stretch  when  carrying  a  load  of  7000  Ibs.     The  tube  is  8'  6"  in. 
long.     What  is  the  factor  of  safety  if  the  ultimate  tensile  resistance  is  50,000 
Ibs.  per  square  inch?    #  =  30,000,000. 

2.  A  wrought  iron  bar  1  in.  in  diameter  and  10  ft.  long  is  .015  in.  longer 
when  loaded  than  when  under  no  stress.     #  =  29,000,000.     Determine  (a) 
the  load  on  the  bar,  (6)  the  stress  per  square  inch,  and,  (c)  the  factor  of  safety 
if  the  ultimate  stress  is  40,000  Ibs.  per  square  inch. 

3.  Sketch  shows  a  hollow  cylindrical  ring  supporting  a  load  of  25  tons. 
Assuming  it  to  be  of  cast  iron  having  an  ultimate  compressive  strength  of 
80,000  Ibs.  per  square  inch,  what  should  thickness  be  if  a  factor  of  safety 
of  20  be  used? 

4.  Find  diameter  D  of  tension  rod  to  sustain  a  load  P  =  40,000  Ibs.,  so 
that  the  stress  in  rod  will  not  exceed  10,000  Ibs.  per  square  inch.     The  pin 
joining  the  two  rods  is  1\  in.  in  diameter,  and  has  a  l^-in.  hole  through  it. 
Determine  the  shearing  stress  in  pin. 

25"  Tons 


-7"- 


- 


P.    . 

CHAP.  I.     Prob.  3.        CHAP.  I.     Prob.  4. 


CHAP.  I.     Prob.  5. 


5.  Sketch  represents  a  punch  and  die.     If  d  =  l  in.,  t  =  \  in.  and  shearing 
strength  of  plate  to  be  punched  is  40,000  Ibs.  per  square  inch  what  force  P 
is  required  to  punch  the  hole? 

6.  A  beam  of  20-ft.  span  is  loaded  at  four  points  equidistant  from  each  other 
and  from  the  two  end  supports,  with  four  equal  loads  of  2  tons  each.     Find 
(a)  the  two  reactions  at  the  supports,  (6)  the  bending  moments  under  each 
load. 

7.  Beam  is  loaded  as  shown  by  two  concentrated  loads,  PI  =2500  Ibs. 
and  P?  =  1500  Ibs.     Find  both  reactions  and  the  bending  moments  under 


18 


ELEMENTS  OF  MACHINE  DESIGN 


each  load.     The  beam  has  a  square  section:   find  its  dimensions  if  the  safe 
stress  is  1000  Ibs.  per  square  inch. 

8.  A  cantilever  beam  is  10-ft.  span.  It  is  loaded  with  a  uniformly  dis- 
tributed load  of  350  Ibs.  per  running  foot  and  a  concentrated  load  of  4000 
Ibs.,  8  ft.  from  support.  Find  maximum  bending  moment;  find  dimensions 
of  section  if  this  is  a  rectangle  the  depth  of  which  is  three  times  its  width. 
The  safe  stress  is  1200  Ibs.  per  square  inch. 


_ 

-5 — * 6 


I 


-18- 


CHAP,  I.    Prob,  7. 


12bia. 


CHAP.  I.    Prob.  10. 


SECTION  A-B 


CHAP.  I.     Prob.  9. 


CHAP.  I.     Prob.  11. 


9.  The  three  cantilevers  A,  B,  and  C  are  of  equal  cross-sectional  area. 
Assuming  same  material  find  ratios  of  their  safe  loads. 

10.  The  bracket  carries  a  load  P  =  10,000  Ibs.     The  section  is  as  shown. 
Determine  the  stress  in  this  section. 

11.  This  bracket  has  section  at  support  as  shown.     Determine  the  load 
which  it  can  carry  if  the  allowable   stress  is  6000  Ibs.  per  square  inch- 


CHAPTER  II 
SCREW    FASTENINGS 

Where  it  is  necessary  to  fasten  two  or  more  parts  together 
in  such  a  way  that  they  may  be  readily  separated  some  form  of  the 
screw  or  bolt  is  most  commonly  used.  The  thread  is  formed  by 
cutting  or  rolling  a  helical  groove  into  the  blank,  the  form  of 
the  cross-section  of  this  groove  being  in  general  either  triangular 
or  rectangular  with  the  corners  sharp  or  rounded.  The  triangular 
thread,  which  is  the  stronger,  is  used  for  screws  and  bolts  the 
purpose  of  which  is  to  fasten  machine  parts  together.  The  rect- 
angular thread,  producing  less  friction,  is  used  where  the  object 
of  the  thread  is  the  transmission  of  power  or  motion. 


FIG.  2-1. 


U.  S.  Standard  Thread.  Fig.  2-1  shows  the  form  of  the 
U.  S.  Standard  or  Sellers  thread.  This  is  an  equilateral  triangle 
with  top  and  bottom  cut  off.  The  pitch,  p,  is  the  distance  from 
one  turn  of  the  thread  to  the  next  turn.  The  amount  cut  off  at  top 

TT 

and  bottom  is  c  =  —  so  that  h  =  %H  =  .65p.     If  D  is  the  outside 

o 

or   nominal   diameter   then   diameter   at   bottom   of  thread   is 
d=D—  1.3p.     The  number  of  threads  per  inch  is  n  =  -.    Table 

1  below  gives  standard  dimensions  of  bolts  as  used  in  this  country. 

The  Whitworth  Thread.     This  thread  is  used  universally  in 

Great  Britain  and  very  largely  on  the  Continent.     The  sides  of 

19 


20 


ELEMENTS  OF   MACHINE  DESIGN 


TABLE  1 

SCREW-THREADS 


Diameter  of 
Screw. 

Number  of 
Threads  per 
Inch. 

Diameter  at 
Bottom  of 
Threads. 

Area  at  Bottom 
of  Threads  in 
Square  Inches. 

Area  of  Bolt 
Body  in 
Square  Inches. 

1 

4 

20 

.185 

.027 

.049 

A 

18 

.240 

.045 

.077 

! 

16 

.294. 

.068 

.110 

A 

14 

.344 

.093 

.150 

\ 

13 

.400 

.126 

.196 

A 

12 

.454 

.162 

.249 

I 

11 

.507 

.202 

.307 

1 

10 

.620 

,  .302 

.442 

1 

9 

.731 

.420 

.601 

i 

8 

.837 

.550 

.785 

H 

7 

.940 

.694 

.994 

H 

7 

.065 

.893 

1.227 

if 

6 

.160 

1.057 

1.485 

1* 

6 

.284 

1.295 

1.767 

if 

5£ 

.389 

1.515 

2.074 

H 

5 

.491 

1.746 

2.405 

U 

5 

.616 

2.051 

2.761 

2 

4| 

1.712 

2.302 

3.142 

2J 

4£ 

1.962 

3.023 

3.976 

2| 

4 

2.176 

3.719 

4.909 

2| 

4 

2.426 

4.620 

5.940 

3 

3i 

2.629 

5.428 

7.069 

3i 

3£ 

2.879 

6.510 

8.296 

3| 

31 

3.100 

7.548 

9.621 

3| 

3 

3.317 

8.641 

11.045 

4 

3 

3.567 

9.963 

12.566 

41 

21 

3.798 

11.329 

14.186 

4i 

2! 

4.028 

12.753 

15.904 

4f 

2| 

4.256 

14.226 

17.721 

5 

2| 

4.480 

15.763 

19.635 

51 

2£ 

4.730 

17.572 

21.648 

5£ 

2f 

4.953 

19.267 

23.758 

5| 

2| 

5.203 

21.262 

25.967 

6 

21 

5.423 

23.098 

28.274 

SCREW  FASTENINGS 


21 


the  thread  make  an  angle  of  55°  with  each  other.  The  top 
and  bottom  are  rounded  off  and  an  amount  c  =  £  H  is  cut  off  from 
the  primitive  triangle.  For  this  thread  d  =  D  —  1.28p. 


FIG.  2-2. 


FIG.  2-3. 


Other  Forms  of  Threads.  The  sharp  V  thread  is  shown  in 
Fig.  2-3.  The  sides  make  an  angle  of  60°  with  each  other.  The 
thread  is  used  both  in  America  and  Europe.  Since  the  section 


FIG.  2-4. 

is  an  equilateral  triangle  d=D—1.732p.  On  account  of  greater 
depth  and  sharp  corner  at  bottom  a  bolt  with  this  thread  is  much 
weakened.  Fig.  2-4  shows  the  square  thread  used  generally 


FIG.  2-5. 


where  the  transmission  of  motion  is  the  purpose  of  the  screw 
is,  for  instance,  the  lead  screw  of  a  lathe.    With  the  proportions 


22 


ELEMENTS  OF   MACHINE  DESIGN 


shown  d  =  D  —  p.     A  modification  of  this  thread  known  as  the 
"  acme  thread  "  is  shown  in  Fig.  2-5.     The  angle  a  =  14j0. 


FIG.  2-7. 


FIG.  2-8. 


The  "  buttress  thread,"  Fig.  2-6,  is  used  where  the  pressure 
always  acts  in  one  direction,  against  the  perpendicular  side,  as 
in  the  breech  mechanism  of  modern  ordnance. 


FIG.  2-9. 


FIG.  2-10. 


The  "  knuckle  thread,"  Fig.  2-7,  formed  by  rounding  the  top 
and  bottom  of  the  square  thread  is  especially  Adapted  for  rough 
usage  a§  it  is  not  liable  to  be  easily  injured 


SCREW  FASTENINGS 


23 


Types  of  Bolts.  Bolts  may  be  divided  into  three  general 
classes:  (a)  through  bolts,  (6)  tap  bolts  or  cap  screws  and  (c) 
stud  bolts.  In  fastening  two  parts  together  by  a  through  bolt 


FIG.  2-11. 


FIG.  2-12. 


(Fig.  2-8)  the  holes  are  "  drilled  "  an  easy  fit  for  the  bolt.  With 
tap  bolts  one  hole  is  drilled  and  the  other  tapped  or  threaded  as 
shown  in  Fig.  2-9.  The  stud  bolt  is  threaded  at  both  ends  as 


FIG.  2-13. 


FIG.  2-14. 


shown  in  Fig.  2-10,  one  hole  being  drilled  and  the  other  tapped. 
A  nut  is  used  to  hold  the  parts  together.  These  bolts  are  made 
in  sizes  from  J  in.  up,  as  shown  in  Table  1. 


24 


ELEMENTS  OF  MACHINE  DESIGN 


FIG.  2-15. 


Forms  of  Boltheads.     Bolts  are  made  with  various  forms  of 

heads  depending  upon  the  use  to  which  they  are  to  be  put.  The 
commonest  forms  being  the  square  and 
hexagon  heads  shown  in  Figs.  2-11  and 
2-12.  A  few  of  the  other  forms  frequently 
met  with  are  shown  in  Figs.  2-13  to  2-16. 
Fig.  2-13  is  the  Tee  head,  Fig.  2-14  is  an 
eye  bolt,  Fig.  2-15  is  a  hook  bolt  and  Fig. 
2-16  is  the  round-head  bolt. 

Machine  Screws.  The  term,  machine 
screw,  is  applied  to  a  variety  of  small 
screws,  generally  with  screw-driver  heads, 
ranging  in  diameter  from  ^  to  |  in.  No 
standard  shave  been  adopted  for  these  up 
to  the  present  time.  The  usual  forms  of 

heads  are  shown  in  Figs.  1-17  to  1-19. 

Set  screws  are  used  to  prevent  relative  rotation  of  two  parts 

as  a  pulley  and  shaft,  the  holding  power  being  due  to  friction 

produced  by  pressure  on  end  of  screw.     In 

Fig.  2-20  is  shown  the  usual  form  of  these; 

a  being  the  flat  end,  b  the  cone,  and  c  the 

cup  end  screw. 

Locking  Device.     When  a  bolt  is  used 

where   it   is   subjected   to   vibration  it  is 

necessary  to  lock  the  nut  to  prevent  it  from 

loosening.     The  commonest  form  of  locking 

device  is  two  nuts,  the  lower  one,  which  is 

the  lock  or  check  nut;  should  be  tightened 

against   the  upper  nut  (Fig.  2-21).      The 

thickness  of  check  nut  is  generally  about 

half    that    of    the    regular    nut.      Other 

methods    of    locking    nuts   are   shown   in 

Figs.  2-22  to  2-24.    In  Fig.  2-23  the  nut  has 

a  cylindrical  extension  which  extends  into 

the  part  to  be  held  by  bolt,  a  set  screw  grips 

this  extension.     In  Fig.  2-24  a  washer  is 

placed  under  the  nut,  a  portion  of  this  washer  is  bent  up  to  fit 

against  one  face,  while  another  portion  is  bent  down  into  an 

opening  thus  preventing  washer  and  nut  fom  turning. 


FIG.  2-16. 


SCREW  FASTENINGS 


25 


FIG.  2-17.        FIG.  2-18.        FIG.  2-19. 


FIG.  2-21. 


V 

FIG.  2-20. 


FIG.  2-22. 


FIG.  2-23. 


FIG.  2-24. 


Pipe  Thread.     The  standard  form  of  pipe  thread  is  the  Briggs 
thread  shown  in  Fig.  2-25.    The  number  of  threads  per  inch  are 

J-in.  pipe 27  threads  per  inch. 

J  and  f-in.  pipe 18       " 

£  and  f-in.  pipe 14       " 

1  to  2-in.  pipe llj     "  " 

2J  ins.  and  over 8       "  " 


'26  ELEMENTS  OF   MACHINE  DESIGN 

Multiple  Threaded  Bolts.  A  bolt  may  have  two  or  more 
threads  cut  upon  it.  The  distance  between  adjacent  turns  is 
again  the  pitch,  p,  while  the  distance  which  one  thread  advances 
in  a  complete  revolution  is  usually  called  the  lead,  L.  Figs. 
2-26  and  2-27  illustrate  a  double  and  triple  threaded  bolt. 
These  bolts  are  used  for  power  transmission  purposes  as  they 
give  a  better  efficiency  than  single  threaded  bolts. 

The  Strength  of  Bolts.  For  purposes  of  design  bolts  may  be 
divided  into  three  classes: 

1.  Those  in  which  the  stress  is  due  to  the  load  only. 

2.  Those  which  are  under  an  initial  stress  due  to  tightening. 

3.  Those  which  are  used  to  transmit  power  or  motion. 

Fig.  1-28  shows  a  bolt  in  which  the  stress  is  that  due  to  the 
load  only.  If  d  =  root  diameter,  P  is  the  tensile  load  and  st  is 
the  safe  tensile  stress,  then 


and 

~   \7Tsi~      '         \  St' 

The  outside  diameter  D  for  the  U.  S.  standard  thread  may  then 
be  obtained  from   Table  1. 

The  second  case  is  illustrated  by  Fig.  2-29  which  shows  one 
of  the  stud  bolts  holding  the  cylinder  head  of  a  steam  engine. 
Here  it  is  necessary  to  screw  down  the  nut  sufficiently  to  make  a 
steam  tight  joint.  The  amount  of  this  initial  tightening  depends 
on  the  relative  elasticities  of  the  material  of  bolts,  flanges  and 
packing. 

Let   P  =  pressure  to  be  sustained  by  each  bolt, 

k  =  a  coefficient  depending  on  kind  of  packing, 

st  =  safe  tensile  stress  in  bolt, 

A  =  root  area  of  bolt, 

d=root  diameter  bolt. 
Then 


(2) 


SCREW  FASTENINGS 


2? 


The  following  values  of  k  may  be  used:  For  a  ground  joint  or 
metallic  packing  such  as  a  copper  ring  fc=lf,  for  soft  packing 
such  as  paper,  asbestos  or  rubber  &  =  lf. 


FIG.  2-25. 


FIG.  2-26. 


FIG.  2-27.T 


EXAMPLE.  Determine  the  diameter  of  bolts  necessary  to  hold 
the  cylinder  head  of  an  engine  having  a  cylinder  diameter  of  10 
ins.  and  using  steam  at  150  Ibs.  per  square  inch.  \Ve  will  assume 


28  ELEMENTS  OF  MACHINE  DESIGN 

that  8  bolts  are  used  and  that  the  gasket  is  of  rubber.  The  safe 
stress  for  these  bolts  should  be  assumed  low  from  4000  to  6000 
Ibs.  per  square  inch;  as  this  is  a  case  of  repeated  or  continually 
varying  loads. 

Total  pressure  on  cylinder  head  is 

W=^X  100X150  =11,800  Ibs. 
Pressure  on  each  bolt  is 

P=^=  1475  Ibs. 

o 

Then  assuming  st  =  6000  we  have    . 


=  .715  in. 
This  requires  a  bolt  %  in.  outside  diameter. 

Transmission  Screws  are  used  for  the  transmission  of  power 
in  hoisting  machines,  screw  presses,  machine  tools,  such  as 
planers  and  Blotters,  and  many  others.  The  power  may  be 
applied  either  to  the  nut  or  the  screw.  The  stresses  induced  in 
screw  are  tension,  or  compression,  combined  with  torsion.  In 
Fig.  2-30  let 

r\  =  outside  radius  of  thread, 
r2  =  inside  radius  of  thread, 

r  =  mean  radius  of  thread, 

p  =  pitch  of  thread, 

/t  =  coefficient  of  friction  between  nut  and  thread, 
//  =  coefficient  of  friction  between  collar  and  support, 
R'  =  mean  radius  of  collar, 

*D 

a  =  angle  of  thread  =  tan"1^—  , 


4>  =  friction  angle  =  tan-  V, 

P  =  force  applied  at  end  of  lever, 

N  =  normal  pressure  between  nut  and  screw  thread, 

TF=load  on  thread. 


SCREW  FASTENINGS 


29 


Since  the  work  done  by  force  applied  at  end  of  lever  must  be 
equal  to  the  work  required  to  raise  the  load  W,  and  to  overcome 
friction  of  nut  and  of  collar,  we  have  for  one  turn  of  the  lever, 


2irRP  = 

N=W  COS  a, 


FIG.  2-29. 


FIG.  2-30. 
The  efficiency  of  the  screw  is 

useful  work  done        Wp 
total  energy  applied     2irRP 


e  = 


P 


If 


p+  2ir(nr  cos  a 
//  and  r  =  Rr  this  reduces  to 
tan  a 


e  = 


tan  a+tan  <£(cos  a-f  1)' 


(3) 


(4) 


The  efficiency  therefore  increases  as  a  increases,  and  it  is  for  this 
reason  that  multiple  threaded  screws  are  used  for  power  trans- 
mission. If  tan  a  is  less  than  //  the  screw  is  self-locking,  that  is, 
no  matter  how  great  the  load  W  may  be,  it  cannot  cause  rotation 
of  the  screw  and  consequent  running  down  of  nut.  This  is  an 
important  consideration  in  the  design  of  transmission  screws  and 
worm  gears  for  hoisting  machinery.  If  tan  a  is  greater  than  the 


30 


ELEMENTS  OF  MACHINE  DESIGN 


above  value  a  force  or  load  applied  to  nut  may  cause  rotation  of 
screw.  A  familiar  example  of  this  is  the  spiral  screwdriver  and 
drill. 

In  calculating  the  diameter  of  a  transmission  screw  it  is  suf- 
ficient to  make  the  root  diameter  safe  to  resist  the  direct  tension 
or  compression  as  the  increase  in  strength  due  to  thread  will  take 
care  of  the  torsion  stresses. 

EXAMPLE.  Design  the  screw  for  a  hoisting  jack  to  raise  a  load 
of  10  tons.  Assume  a  safe  stress  of  8000  Ibs.  per  square  inch, 
then  the  root  diameter  is 


=  1.79  in. 

If  a  pitch  of  |  in.  be  assumed  then  the  outside  diameter  of  the 
square  threaded  screw  is 

d+p  =  1.79+i  =  2.39  ins.,  say  2J  ins. 
The  coefficient  of  friction  may  be  assumed  at  .10  then 
0  =  tan-1.10=5°40', 


tan 


-i. 


.5 


=  4°. 


27T1.25 

The  efficiency  of  the  screw  therefore  is 

tan  4° 


e  = 


tan  4°+tan  5° -40' (cos  4° +1) 
.07 


.07+.10(.998+1) 


=  26  per  cent. 


It  should  be  noted  that,  the  screw  is  subjected  to  a  compres- 
sive  load,  it  may  have  to  be  designed  as  a 
column  if  it  be  long  in  comparison  to  its 
diameter.  At  any  rate  it  will  be  well  to 
check  the  safe  load  according  to  the  column 
equation  of  Chapter  I. 

If  the  load  on  a  bolt  is  not  applied 
centrally  with  the  bolt  there  will  be  pro- 
duced a  bending  stress  besides  the  direct 
stress.  Thus  in  Fig.  2-31  the  bolt  is  sub- 


FIG.  2-31. 


SCREW   FASTENINGS  31 

jected  to  a    combined    direct  tension  P  and  a  bending  moment 
equal  to  PL 

PROBLEMS 

1.  In  a  screw  press  a  force  of  25  tons  is  transmitted  to  the  platen  by 
means  of  two  square  threaded  screws  of  6  threads  to  the  inch.      Determine 
outside  diameter  of  these  screws  if  a  tensile  stress  of  8000  Ibs.  be  allowed. 

2.  The  cylinder  head  of  a  steam  engine  is  held  by  10  stud  bolts.     The 
diameter  of  cylinders  is  12  ins.  and  the  steam  pressure  is  125  Ibs.  per  square 
inch.      Find  root  diameter  of  bolts  if  safe  stress  is  4000  Ibs.  per  square  inch. 

3.  In  a  turnbuckle  the  threaded  ends  are  1^  ins.  outside  diameter,  4  threads 
per  inch.      The  material  is  soft  steel  having  a  tensile  strength  of  40,000  Ibs. 
per  square  inch.     The  turnbuckle  was  tightened  until  rupture  of  one  of  the 
ends  occurred.     What  was  the  load  on  the  rods? 

4.  A  jack  screw  is  to  be  designed  to  raise  a  maximum  load  of  30  tons. 
Determine  its  root  diameter  if  a  factor  of  safety  of  4  be  employed  and  the 
material  is  mild  steel. 

5.  In  a  hydraulic  press  the  upper  platen  is  held  by  four  bolts.     The 
pressure  is  exerted  by  a  plunger  15  ins.  diameter  with  a  maximum  hydraulic 
pressure  of  400  Ibs.  per  square  inch.     Determine  size  of  bolts  if  the  safe  stress 
is  8000  Ibs.  per  square  inch. 

6.  What  is  the  efficiency  of  a  square  threaded  screw  3  ins.  diameter.     It 
is  triple  threaded,  the  lead  being  1  in.     Assume  coefficient  of  friction  is  .12. 


CHAPTER  lit 
RIVETED  JOINTS 

Types  of  Rivets  and  Joints.  When  two  parts  are  to  be  per- 
manently fastened  together  the  riveted  joint  is  commonly  used. 
The  rivet  consists  of  a  head  and  a  cylindrical  shank  slightly 
tapered  at  the  end  (Fig.  3-1).  It  is  heated  to  a  bright  red  heat 


FIG.  3-1. 


FIG.  3-2. 


and  another  head  called  "  point  "  is  formed,  either  by  pressure 
or  by  hammering,  as  shown  in  Fig.  3-2.  Riveted  joints  are  used 
chiefly  in  structural  steel  work  and  for  connecting  the  plates  of 
vessels  under  pressure  such  as  boilers  and  tanks.  In  structural 
work  strength  of  joint  is  the  chief  consideration  while  in  boiler 


H— K»D 


CONE  HEAD  BUTTON  HD*         ^TEEPLE  HD. 

FIG.  3-3. 


COUN'RSUNK  HEAD 


joints  and  similar  work  the  tightness  of  joint  to  prevent  leakage 
is  of  equal  importance.  Fig.  3-3  shows  the  various  forms  of  rivet 
heads  which  are  commonly  used. 

Boiler  joints  are  of  two  general  types  called  lap  joints  and 
butt  joints.     In  the  lap  joints  the  plates  to  be  riveted  together 

32 


RIVETED  JOINTS 


33 


overlap  as  shown  in  Fig.  3-4.  Depending  on  the  number  of  rows 
of  rivets  the  joints  is  single,  double,  triple  or  quadruple  riveted. 
In  the  butt  joint  shown  in  Fig.  3-5  the  ends  of  the  plates  butt 


ojo 

olo 

o;o 

oj© 

FIG.  3-4. 


FIG.  3-5. 


against  each  other  and  a  strip  of  metal,  called  a  butt  strap  or 
welt,  is  placed  on  one  or  both  sides  of  the  plates.  A  single  riveted 
butt  joint  has  one  row  of  rivets  on  each  side  of  joint,  a  double 
riveted  butt  joint  has  two  rows,  and  so  on.  The  butt  joint  is 


FIG.  3-6. 


considered  the  safest  and  is  generally  used  in  high-pressure  boilers, 
Figs.  3-4  to  3-10  show  some  of  the  common  forms  of  boiler 
joints. 

The  material  used  for  rivets,   plates  and  structural  shapes 
is  usually  open-hearth  or  Bessemer  steel,  wrought  iron  being  but 


34 


ELEMENTS  OF  MACHINE  DESIGN 


rarely  used  in  present  day  practice.  For  boiler  work  open- 
hearth  steel  is  used  almost  exclusively,  there  being  three  grades, 
called  flange  or  boiler  steel,  fire-box  steel  and  extra  soft  steel. 
Their  tensile  and  shearing  strengths  range  as  follows: 


Tensile  Strength. 

Shearing  Strength. 

Flange  st6cl 

55,000-65,000 

48000 

Fire-box  steel       

52,000-62,000 

48,000 

Extra  soft  steel            

45,000-55,000 

45,000 

Rivets  are  made  of  bars  of  extra  soft  steel. 


o  o  o  o* 

1 

r 

';' 

,  , 

L_ 

) 

x  —  S            ^-/            \~**s 
O/~N           /^N           S^. 

I 

' 

x 

^s. 

O     O    (Q) 

)l 

O/—  N           X—  \          ^-N 

y 

O    (J   C3 

) 

,;/ 

y 

A 

L 

FIG.  3-7. 


Calking.  Joints  are  made  fluid  tight  by  calking  as  shown  in 
Fig.  3-11.  For  this  purpose  the  edges  of  the  plates  to  be  calked 
are  planed  off  at  an  angle  of  about  80°.  These  edges  are  then 
burred  down  by  means  of  the  calking  chisel.  This  work  is  done 
either  by  hand  or  pneumatic  hammer.  Skill  and  care  are  required 
to  prevent  injury  of  the  plate. 

Strength  of  Riveted  Joints.  The  distance  between  adjacent 
rivets  in  the  same  row  is  called  the  pitch  of  the  rivets  and  will 
be  indicated  by  p.  If  the  pitch  of  the  rivets  in  the  different 
rows  is  not  the  same,  the  maximum  pitch  will  be  understood  by 
p,  as  in  Fig.  3-9.  When  the  rivets  in  the  various  rows  are 
opposite  each  other  it  is  galled  chain  riveting  (Fig.  3-7).  If 


HIVETED  JOINTS 


35 


they  are  staggered  it  is  called  zigzag  riveting  (Fig.  3-6).     The 
distance  between  center  lines  of  adjacent  rows  of  rivets  is  the 


u>   u>   LJ    o 

( 

) 

^  —  S               x  '              X.  f              \,  ' 

O  O  O 

\ 

c 

- 

) 

o  6  of 

O^\        ^^       s~\ 

/ 

c 

X" 

- 

.: 

) 

)"" 

o  o  © 

v»  —  '            ^^~/            v._./ 

... 

.. 

FIG.  3-8. 


O 


o  o 
o  o 


o 
o  o 


o 


C 


c 


FIG.  3-9. 


transverse  pitch  and  will  be  indicated  by  ph  while  the  diagonal 
pitch,  pa,  is  the  distance  from  the  center  of  one  rivet  to  that  of 
the  nearest  to  it,  diagonally,  in  the  next  row  (Fig.  3-6). 


36 


ELEMENTS  OF  MACHINE  DESIGN 


Failure  of  a  joint  may  occur  in  any  one  of  the  following  ways; 

1.  Tearing  of  plate  between  rivets  as  in  Fig.  3-12. 

2.  Shearing  of  rivets  as  in  Fig.  3-13. 

3.  Crushing  of  plate  or  rivet  as  in  Fig.  3-14. 

4.  Tearing  of  plate  in  front  of  rivet  as  in  Fig.  3-15. 

5.  Shearing  of  plate  in  front  of  rivet  as  in  Fig.  3-16. 


o  o  o  o 


o  o  o  o 


o  o  o  o 
o  o  o  o 
o 


FIG.  3-10. 

In  the  more  complex  types  of  joints  failure  may  occur  by  a  com- 
bination of  two  or  more  of  the  above  causes. 

The  ratio  of  the  strength  of  weakest  element  of  the  joint 
to  that  of  the  solid  plate  is  called  the  efficiency  of  the  joint.  It 
is  usually  expressed  in  per  cent.  If  the  centers  of  rivets  be  placed 
from  one  and  one-half  to  twice  the  rivet  diameter  from  the  edge 
of  plate,  failure  by  methods  4  and  5  will  not  occur  and  therefore 


RIVETED  JOINTS 


37 


consideration  of  these  methods  of  failure  will  be  omitted  from 
the  following  discussion. 

Let    d  =  diameter  of  rivet  in  inches, 
t  =  thickness  of  plate  in  inches, 
Ut  =  tensile  resistance  of  plate, 
Uc  =  crushing  resistance  of  plate  or  rivet, 
Us  =  shearing  resistance  of  rivet  in  single  shear. 


FIG.  3-11. 


FIG.  3-12. 


FIG.  3-13. 


FIG.  3-14. 


FIG.  3-15. 


FIG.  3-16. 


The  shearing  resistance  of  a  rivet  in  double  shear  will  be  taken 
at  1.8  the  shearing  resistance  of  the  same  rivet  in  single  shear. 
Since  portions  of  the  joint  of  a  length  equal  to  the  pitch  are 
identical  it  is  only  necessary  to  consider  such  a  part  of  the  joint 
and  it  will  be  called  a  unit  strip. 

Let  R  =  tensile  resistance  of  a  unit  strip  of  the  solid  plate, 
Rt  =  tensile  resistance  of  a  unit  strip  of  the  joint, 
Rs  =  shearing  resistance  of  a  unit  strip  of  the  joint, 
Rc  =  crushing  resistance  of  a  unit  strip  of  the  joint, 

r» 

Et  =  tensile  efficiency  of  the  joint  =  WX  100  per  cent, 


38  ELEMENTS  OF   MACHINE   DESIGN 

r> 

Es  =  shearing  efficiency  of  the  joint  =-^X  100  per  cent, 

r> 

Ec  =  crushing  efficiency  of  the  joint  =  ~^X  100  per  cent. 

tii 

Single  Riveted  Lap  Joint  (Fig.  3-4).  In  this  joint  there  is 
one  rivet  in  the  unit  strip.  The  cross-sectional  area  of  a  solid 
strip  of  the  plate  having  a  width  equal  to  p  and  thickness  t  is  pt, 
therefore  its  resistance  to  tearing  is 

R  =  ptUt. 

If  we  now  take  a  unit  strip  of  the  joint,  due  to  the  rivet  hole 
the  sectional  area  of  the  plate  is  reduced  by  dt,  therefore  the  re- 
sistance to  tearing  now  is 

Rt=(p-d)tUt. 

Since  in  the  length  of  joint  equal  to  p  there  is  one  rivet  only 
the  shearing  resistance  is 


Although  the  pressure  between  rivet  and  plate  is  distributed 
over  a  semi-cylindrical  area  it  is  customary  to  assume  that  the 
result  is  that  of  a  uniformly  distributed  stress  on  an  area  equal  to 
dt  for  each  rivet,  which  is  the  projected  area  of  the  cylindrical 
surface,  then 

Rc  =  dtUc. 
From  the  above  equations  we  obtain  the  efficiencies 


R  p 


RIVETED  JOINTS  39 

Double-riveted  Lap  Joint  (Fig.  3-6).  In  this  joint  there 
are  two  rivets  in  the  unit  strip  and  the  following  equations 
may  readily  be  established  : 

R  =  ptUt, 

Rt=(p-d)tUt, 


Rc  =  2dtUc. 

The  efficiencies  obtained  in  the  same  manner  as  for  single 
riveted  lap  joint  are: 


~2ptUt 

F_2dUc 
c~~pU~t> 

Triple-riveted  Lap  Joint.     In  this  joint  (Fig.  3-7)  there  are 
are  3  rivets  in  the  unit  strip.    Proceeding  as  before  we  obtain 

R=ptU,, 

Rt=(p-d)tUh 


Rc=ZdtUc. 

The  efficiencies  are 


40  ELEMENTS  OF  MACHINE  DESIGN 

Another  form  of  triple-riveted  lap  joint  has  the  pitch  of  the 
outer  rows  of  rivets  twice  that  of  the  inner  row  (Fig.  3-7  a).  In 
this  joint  there  are  four  rivets  in  the  unit  strip: 

R  =  ptUt, 

Rt=(p-d)tUt, 


Rc=4dtUe. 

Another  method  of  failure  is  possible  with  this  joint,  viz.: 
a  combination  of  tearing  of  plate  and  shearing  of.  rivets.  Thus  the 
plate  may  tear  at  the  center  row  of  rivets  and  before  failure 
could  occur  one  of  the  outer  rows  of  rivets  would  have  to  be 
sheared.  For  this  case 

Rts=(p-2d)tUt+^d2Us. 
The  efficiencies  therefore  are 


,P_-2d  «x 

\     p        4ptUt/ 


*-s*l* 

« 

EXAMPLE.  What  is  the  efficiency  of  a  double-riveted  lap  joint, 
diameter  of  rivets  f  in.,  thickness  of  plate  f  in.,  pitch  2f  in.? 
Rivets  and  plates  are  of  steel  and  t/,  =  60,000,  Us  =  45,000, 
Uc  =  80,000. 

=  70  per  cent, 


irX.752X45000 


2X.75X80000XX 
=    2.5X60000    X100-80  per  cent. 


Actual  efficiency  is  therefore  70  per  cent. 


RIVETED  JOINTS  41 

Single-riveted  Butt  Joint  (Fig.  3-5).  In  this  joint  the  rivets 
are  in  double  shear.  There  is  one  rivet  on  each  side  of  the  joint 
in  the  unit  strip. 

R=ptUh 

Rt  =  (p-d)tUt) 


c  =  dtUc 


,  „ 

4ptUt  ™> 


Double-riveted  Butt  Joint  (Fig.  3-8).  In  this  joint  there  an 
two  rivets  in  double  shear  in  the  unit  strip.  All  rivets  are  in 
double  shear. 

R  =  ptUt, 

Rt  =  (p-d)Ut, 


=  2dtUc 


2dUc 
tU, 


42  ELEMENTS  OF  MACHINE  DESIGN 

Triple-riveted  Butt  Joint  (Fig.  3-9).     In  this  joint  the  pitch 
of  the  outer  rows  of  rivets  is  twice  that  of  the  inner  rows. 

R  =  ptUt) 
Rt=(p-d)tUt, 


Another  method   of  failure  possible  is  tearing  of  plate  at 
middle  row  and  shearing  of  outer  row  of  rivets.    For  this  case 


The  efficiencies  are 


2.Qs 

~~    °°' 


?c=^xioo, 

/v  —  2d 
&•=(- 

V    p 


EXAMPLE.  Find  the  efficiency  of  a  triple-riveted  butt  joint, 
Fig.  3-9.  The  pitch  p  is  1\  ins.,  thickness  of  plate  |  in.,  and  diam- 
eter of  rivet  holes  1  in. 

=  87  per  cent; 


_2.057r^2?7sXlOO_2.05X7rXlX4500QXlOO_1ono/ 
p+Ut  7.5X.  5X60000 

5X1X80000   _ 
7.5X.5X60000" 

7.5-2  7TX45000 


The  actual  efficiency  of  the  joint  is  therefore  87  per  cent. 


RIVETED  JOINTS  43 

Design  of  Riveted  Joints.  From  the  preceding  it  is  evident 
that  the  efficiency  of  a  joint  depends  on  three  variables,  thickness 
of  plate  t,  diameter  of  rivets  d,  and  the  pitch  p.  Practice  has 
fixed  the  size  of  the  rivets  to  be  used  with  any  thickness  of  plate 
and  the  following  equation  will  give  good  results  for  boiler  joints: 


-Q   n. 

In  designing  a  riveted  joint  it  is  usual  to  make  its  resistance 
to  tension  equal  to  its  shearing  resistance,  or 

Rt  =  Rs* 

From  this  equation  we  may  obtain  the  pitch  for  any  type  of  joint. 
Thus  for  a  single-riveted  lap  joint 

Rt  =  Rs, 


and 


EXAMPLE.     Design    a    double-riveted    butt   joint    for   plate 
f  in.  in  thickness. 

=  .74  =  f -in.  rivet. 

This  is  the  size  of  hole  in  plate  or  rivet  diameter  after  it  is 
driven. 

Rt  =  Rs, 

1.87Td2Us 

(p-d)tUt  = ^ — , 


=  .97r(f)245000     3 
1X60000  +4' 

-Win. 

However,  in  the  design  of  boiler  and  similar  joints  it  must  be 
kept  in  mind  that  the  maximum  pitch  is  determined  by  the  possi- 
bility of  making  a  tight  joint  and  this  will  often  modify  the  cal- 
culated pitch.  The  minimum  pitch  should  not  be  less  than 
$d  in  order  to  drive  the  rivets  and  form  the  head  readily. 


44 


ELEMENTS  OF  MACHINE  DESIGN 


TABLES  SHOWING  DETAILS  OF  RIVET  LAPS  FOR  DIFFERENT 
THICKNESSES  OF  BOILER  PLATE  AS  ADVOCATED  BY  THE 
HARTFORD  STEAM  BOILER  INSPECTION  AND  INSURANCE 
CO.  FOR 

DOUBLE  RIVETED  BUTT  JOINTS 


9 

.j. 

i 

• 

a 

"55 

TJ 

jj 

§ 

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ro 

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m 

s 

II 

Diamete 
Rivets. 

"o-o 

p 

-O  3 

11 
SS 

Vertical 
verse  P 

•ill 

ll 

^o 

Strength 
Joint. 

In. 

In. 

In. 

In. 

In. 

In. 

In. 

In. 

In. 

In. 

% 

A 

H 

21X4^ 

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TRIPLE  RIVETED  BUTT  JOINTS 


gffl 


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Id 


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ering 


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verse 


si. 


Id 

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goo 


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S'e 


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In. 


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6| 

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91 
9| 
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11 


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HI 

148 

14 

141 

141 

15f 

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The  butt  straps  are  usually  somewhat  thinner  than  the  plates 
to  be  joined.  Generally  their  thickness  is  3^  to  |  in.  less  than 
that  of  the  plates.  In  triple  and  quadruple  riveted  butt  joints 
the  butt  straps  are  often  of  different  widths,  as  shown  in  Fig. 
3-9.  Sometimes,  especially  in  marine  work,  the  edges  are  scal- 
loped, as  shown  in  Fig.  3-10.  Both  these  devices  enable  the  use 
of  a  wide  pitch  in  the  outer  rows  of  rivets  and  at  the  same  time, 
permit  the  calking  of  the  joint, 


RIVETED  JOINTS 


45 


PROBLEMS 

1.  In  a  single- riveted  lap  joint  the  thickness  of  plate  is  f  in.,  the  diameter 
of  rivet  is  f  in.,  and  the  pitch  is  2|  in.     Find  the  efficiency  of  this  joint. 

2.  Design  a  double-riveted  lap  joint  making  shearing  resistance  equal  to 
tearing  resistance.     The  plate  is  £  in.  thick  and  rivets  are  £  in.  diameter. 

3.  In  a  triple-riveted  lap  joint  the  pitch  of  the  inner  rows  of  rivets  is 
2f  in.     The  outer  row  of  rivets  is  of  double  the  pitch.     The  rivets  are  1  in. 
diameter  and  the  plates  are  f  in.  thick.     Find  the  efficiency  of  this  joint. 

4.  Design  a  double-riveted  butt  joint  for  f-in.  plate.     Find  efficiency  of 
joint. 

5.  Sketch  shows  a  method  of  joining  two  plates  frequently  used  in  struc- 
tural work.     Find  resistance  of  joint  to  tearing  at  outer  row  of  rivets,  at  second 
row,  and  resistance  to  shearing  of  rivets.     What  is  efficiency  of  this  joint? 


k 

i 

1                                 %"       { 

t 

wwwwwv^y 

ojo    

•13/i'c  Rive' 

1 

r 

o0o|00o 

CHAP.  III.     Prob.  5. 

6.  Design  a  triple-riveted  butt  joint  of  the  form  shown  in  sketch.     The 
thickness  of  plate  is  \  in.     What  is  the  efficiency  of  this  joint? 
if  ^ 


o 

o 

o 

o 

o 

0 

o 

o 

o 

0 

o 

o 

o 

o 

0 

o 

-0 

_O  j 

—  ^-—  J 

V  I 

CHAP.  III.     Prob.  6. 

7.  In  a  double-riveted  butt  joint  the  outer  row  of  rivets  is  in  single  shear 
and  of  double  the  pitch  of  inner  row.     The  plate  is  f  in.  thick,  the  rivets 
1  in.  diameter  and  pitch  of  outer  row  is  5  in.     Determine  efficiency  of  joint. 

8.  Design  a  triple-riveted  butt  joint  for  f-in.  plate.     All  rivets  are  in  double 
shear  and  of  same  pitch.     Find  efficiency  of  this  joint. 


CHAPTER  IV 
KEYS  AND  COTTERS 

Kinds  of  Keys.  Keys  are  rectangular  or  round  bars,  the  pur- 
pose of  which  is  to  prevent  relative  rotation  of  two  parts,  such  as 
a  shaft  and  pulley.  The  sunk  key  shown  in  Fig.  4-1  is  the  com- 
monest type.  For  light  work  the  flat  key,  Fig.  4-2,  and  the  saddle 
key,  Fig.  4-3,  are  sometimes  used.  When  the  hub  is  very  light, 


FIG.  4-1. 


FIG.  4-2. 


FIG.  4-3. 


FIG.  4-4. 


FIG.  4-5. 


Taper3/16tol2 


FIG.  4-6. 

round  or  pin  keys  may  be  used  as  in  Fig.  4-4.  For  very  heavy 
work,  such  as  the  belt  wheels  of  steam  engines,  two  tangential 
keys  are  often  used,  as  shown  in  Fig.  4-5.  To  facilitate  removal 
of  keys  they  are  frequently  provided  with  gib  heads,  Fig.  4-6. 
To  obtain  a  tight  fitting  key  the  top  is  tapered  from  J  to  J  in. 

46 


KEYS  AND  COTTERS 


47 


per  foot  of  length.  When  the  key  prevents  relative  rotation 
but  permits  motion  lengthwise  it  is  called  a  feather  key  or  spline. 
In  this  case  the  sides  of  the  key  are  parallel  and  it  is  fastened 
by  screws  or  otherwise  (Fig.  4-7  and  4—8)  to  one  of  the  two 
parts  it  connects  and  is  an  easy  fit  in  the  other. 


FIG.  4-7. 

Stresses  in  Keys.  The  twisting  moment  which  a  key  trans- 
mits produces  shearing  stresses  in  the  plane  CD  (Fig.  4-9)  and 
crushing  on  sides  of  key. 


FIG.  4-8 

Let   T=PR  =  twisting  moment  to  be  transmitted, 
1= length  of  key. 


48 


ELEMENTS  OF  MACHINE  DESIGN 


Since  the  moment  of  the  shearing  stress  and  of  the  compressive 
stress  about  center  of  shaft  must  each  equal  the  twisting  moment 
to  be  transmitted, 


and 


then  equating  (1)  and  (2), 


(1) 


(2) 


and 


FIG.  4-9. 


(3) 


If  a  key  fits  tightly  in  the  key  seat  the  crushing  resistance  is  con- 
siderably increased  and  a  value  of  J  to  }  may  be  assumed  for 

— ,  for  steel  keys:     This  makes  t  =  %b  to  t  =  b;  the  latter  is  a  square 

Sc 

key  which  is  used  by  manufacturers  of  transmission  machinery. 

It  is  usually  assumed  that  the  key  should  be  capable  of  trans- 
mitting the  entire  twisting  moment  of  the  shaft  upon  which  it 
is  placed,  then 


*  This  is  the  twisting  moment  which  can  be  transmitted  by  a  shaft  of 
diameter  =  d,  see  Chapter  VI. 


KEYS  AND  COTTERS 


49 


and 


A  common  value  of  b  is  \d  and  substituting  this  in  (4)  we  obtain 
l=~d  as  the  minimum  length  of  key. 


Manufacturers  have  not  adopted  standard  dimensions  for 
keys,  and  both  square  and  rectangular  keys  are  usecl.  The  fol- 
lowing tables  taken  from  catalogs  of  prominent  transmission 
machinery  firms  give  good  average  practice: 

TABLE   I 

SQUARE  KEYS 


Shaft  Dia. 

Width. 

Thickness. 

Shaft  Dia. 

Width. 

Thickness. 

u 

& 

& 

2* 

A 

A 

H 

A 

A 

3 

H 

H 

u 

A 

& 

3| 

f 

i 

4 

2 

A 

A 

4 

1 

1 

TABLE  II 
RECTANGULAR  KEYS 


Shaft  Dia. 

Width. 

Thickness. 

Shaft  Dia. 

Width. 

Thickness. 

U 

A 

A 

3* 

1 

1 

u 

1 

\ 

4 

1 

f 

H 

A 

A 

5 

n 

H 

2 

i 

A 

6 

u 

H 

2* 

f 

f 

7 

li 

1 

3 

I 

A 

Cotters.  A  cotter  is  a  form  of  key  in  which  the  forces  tending 
to  separate  the  parts  connected  act  at  right  angle  to  their  axis. 
Figs.  4-10  to  4-13  show  some  of  the  common  forms  of  cottered 


50 


ELEMENTS  OF  MACHINE  DESIGN 


joint.     In  Fig.  4-11  a  gib  is  used  with  the  cotter.     This  prevents 
the  ends  from  spreading  when  the   cotter  is  driven.     It  also 


-dr 


w 


Fia.  4-10. 


FIG.  4-11. 


FIG.  4-12. 


simplifies  the  construction  in  avoiding  a  tapered  hole.  Fig. 
4-12  shows  a  cotter  with  double  gib  and  means  for  locking  cotter. 
In  Fig.  4-13  the  cotter  has  a  double  taper  with  a  locking  device, 


KEYS  AND  COTTERS 


51 


The  Strength  of  Joint.     In  Fig.  4-10  let  a  tensile  load  of 
W  Ibs.  be  applied  to  rod,  then  its  diameter  is 


To  make  the  rod  through  the  cotter  (Fig.  4-14)  of  equal  strength 
it  must  have  an  equal  cross-sectional  area,  that  is 


FIG.  4-13. 


FIG.  4-14. 


The  thickness,  t,  of  cotter  may  be  expressed  in  terms  of  cfe,  it 
being  usually  from  \d%  to  \d%,  or  say  kdi,  then 


and 


*.. 


* 


V 'l-  1.27k 

For  the  two  extreme  values  k  =  J  and  k  =  %  this  reduces  to 

to 


(5) 


The  socket  at  its  weakest  point  has  a  section  as  in  Fig.  4-15. 
This  section  is  subjected  to  tension  and  from  equation 


W=Ast 


52  ELEMENTS  OF  MACHINE  DESIGN 

we  obtain 

^-=A=^(ctf-d22)-t(dz-d2)  .....     (6) 

This  equation  may  be  solved  for  d3,  or  better,  the  outside  diameter 
may  be  assumed  and  equation  (6)  solved  for  stress  st. 

The  cotter  is  in  double  shear  as  shown  in  Fig.  4-16,  then 

W=Ass,     ' 


The  cotter  is  also  subjected  to  compression,  the  compressive 
stress  is 


To  prevent  shearing  out  of  the  end  of  rod  and  of  socket  the  cotter 
should  be  placed  in  from  these  ends  a  distance  at  least  equal 
to  %h  and  preferably  J/L 

The  taper  of  the  cotter  is  usually  from  J  in.  in  12  ins.  to  1  in. 
in  12  ins.  To  obtain  a  self-locking 
cotter,  that  is,  one  in  which  the  cotter 
will  not  loosen  no  matter  how  large  the 
force  TF,  the  angle  a  (Fig.  4-10)  must 
be  such  that  tan  <*<M;  where  n  is  the 


FIG.  4-16.  coefficient  of  friction  between  the  sur- 

face   in    contact.      Where   the  joint  is 

subjected  to  a  live  or  variable  load  (connecting  rods)  the  cotter 
is  prevented  from  loosening  by  a  set  screw. 

PROBLEMS 

1.  Find  breadth  of  a  key  which  is  to  transmit  20  H.P.  at  120  R.P.M., 
the  length  of  key  is  3  ins.  The  stress  not  to  exceed  8000  Ibs.  per  square  inch. 
If  thickness  of  key  is  \  of  breadth  plus  f  in.  what  is  the  compression  stress. 
The  diameter  of  shaft  is  2£  in. 


KEYS  AND  COTTERS  53 

2.  A  pulley  is  keyed  to  a  shaft  3  ins.  in  diameter  by  a  key  6  ins.  long. 
Find  breadth  and  thickness  of  key  such  that  strength  of  key  is  equal  to  strength 
of  shaft.     The  stresses  to  be  ss  =  9000  and  sc  =20,000. 

3.  A  lever  40  ins.  long  is  keyed  to  a  shaft  3  ins.  diameter.     A  load  of  1000 
tt?s.  acts  at  end  of  lever.     Design  the  key. 

4.  Design  a  cottered  joint  to  fasten  a  piston  rod  to  the  cross-head.     The 
diameter  of  engine  cylinder  is  30  ins.  and  the  steam  pressure  is  150  Ibs.  per 
square  inch.     Assume  steel   for   all   parts,  making  s^  =  8000  Ibs.,  ss  =  6000. 
Make  thickness  of  cotter  (t).3  of  diameter  of  rod  at  point  where  it  is  placed. 

5.  In  a  hydraulic  press  the  top  platen  is  held  by  four  rods  cottered  into 
base  of  machine.     The  ram  of  press  is  12  in.  diameter  and  the  maximum 
pressure  is  600  Ibs.  per  square  inch.     Design  the  cotter  joint  making  all 
necessary    assumptions. 

6.  Two  steel  rods  2£  ins.  diameter  are  connected  by  means  of  a  cottered 
joint  at  their  ends.     The  socket  formed  at  the  end  of  one  rod  is  6  ins.  out- 
side diameter  and  3  ins.  inside  diameter.     The  cotter  is  f  in.  thick  and  has  a 
depth  of  3|  in.     Determine  how  failure  of  this  joint  would  occur. 


CHAPTER  V 
SHRINK  AND  FORCE  FITS 

When  two  pieces  have  to  be  fastened  together  very  firmly 
shrink  or  force  fits  are  often  employed.  The  elastic  stresses 
produced  in  the  parts  are  the  forces  which  hold  them  together. 
In  shrink  fits  one  of  the  members  to  be  fastened  together  is  heated 
and  in  this  condition  it  fits  over  the  other  member;  upon  cooling 
it  contracts,  thus  gripping  the  inner  one  firmly.  In  force  fits 
the  outer  member  is  made  slightly  smaller  than  the  inner  one  and 
forced  over  it  by  heavy  pressure. 


FIG.  5-1.  FIG.  5-2. 

Shrink  links  are  frequently  used  to  hold  the  halves j)f  a  large 
gear  or  fly-wheel  together.  Their  forms  are  shown^  in  Figs. 
5-1  to  5-5.  Take  the  case  of  link  in  Fig.  5-4  and  let 

A  =  sectional  area, 

1= length  before  heating, 
st  =  tensile  stress  per  square  inch, 
E  =  modulus  of  elasticity, 
X  =  total  elongation  due  to  heating, 
T= total  tension  in  link, 
a  =  coefficient  of  expansion, 

t  =  theoretical  temperature  increase  necessary; 
then 

stress     st  ' 


E= 


strain     V 
I 
54 


SHRINK  AND  FORCE  FITS 


55 


and 


T=Ast=AE£; 


In  the  above  the  compression  produced  in  the  parts  held 
together  has  been  neglected,  which  may  be  safely  done,  as  it  is 
extremely  small  in  all  cases  where  such  links  are  used. 


FIG.  5-3. 


PIG.  6-4. 


FIG.  5-5. 


FIG.  5-6. 


EXAMPLE.  A  crack  in  the  frame  of  a  machine  is  to  be  repaired 
by  means  of  a  shrink  link  as  shown  in  Fig.  5-6.  Assume  steel 
links  and  bolts,  £7  =  30,000,000  and  X  =  .01  in.,  A  =2  sq.in.,  then 

X    30000000  X. 01 


"I  10 

=  30,000  Ibs.  per  square  inch; 


and 


The  bolts  are  in  shear  and  the  stress  is 
T      60000 

S'=^=^T 


=  25,000  Ibs.  per  square  inch. 


56  ELEMENTS  OF  MACHINE  DESIGN 

In  practice  the  grade  of  workmanship  and  finish  of  surfaces 
fitted  greatly  affect  the  shrinkage  allowance  and  the  values  as 
calculated  above  would  be  too  small  for  any  but  the  highest  grade 
of  work  where  accuracy  is  possible. 

Allowance  for  Force  and  Shrink  Fits.  Crank  pins  are  fre- 
quently fitted  into  their  cranks  by  shrink  or  force  fits;  built-up 
crank-shafts  such  as  are  used  for  marine  engines  are  assembled 
in  the  same  way;  car  wheels  are  fitted  to  their  axles  by  force 
fits,  and  the  steel  tires  of  locomotive  wheels  are  shrunk  on.  The 
difference  in  diameters  of  the  two  surfaces  fitted  together  is 
termed  the  allowance.  This  varies  greatly  in  practice  but  it 
may  be  expressed  in  the  form 


Where  x  is  the  allowance  in  thousandths  of  an  inch  per  inch  of 
diameter,  k  is  a  constant  and  d  is  the  diameter  of  the  parts  fitted. 
For  shrink  fits  a  value  of  A:  =4  may  be  used  and  for  force  fits 
k  =  6  gives  results  agreeing  well  with  general  practice.  It  must 
be  understood,  however,  that  the  value  of  k  varies  -with  the 
class  of  work  and  the  finish  and  accuracy  of  surfaces. 

EXAMPLE.     A  crank  web  is  to  be  shrunk  on  a  crank-pin  9  ins. 
diameter;  what  is  the  diameter  of  hole  in  web? 


Total  allowance  =  9X]       =  I        in., 
Diameter  of  hole  =  9  -.012  =  8.988  in. 


CHAPTER  VI 
SHAFTS  AND  AXLES 

A  bar  which  turns  about  its  axis  and  transmits  power  is  termed 
a  shaft.  Formerly  wrought  iron  was  the  favorite  material  used 
for  shafting,  but  on  account  of  lower  cost  and  greater  strength 
steel  is  now  used  almost  exclusively.  For  ordinary  shafting  mild 
steel  is  employed.  Where  the  work  is  very  heavy,  and  weight 
is  to  be  kept  low,  alloy  steels  are  frequently  used.  Thus  in  motor 
cars  and  in  marine  engines  vanadium 
and  nickel  steels  are  used.  Commercial 
shafting  is  either  turned  or  cold-rolled. 
Turned  shafting  is  made  by  turning  down 
bars  of  steel  made  from  the  ingot  by 
hot-rolling.  It  is  xV  in.  less  in  diameter 
than  its  "  nominal  size."  Thus  2-in. 
shafting  is  lr|-in.  diameter.  Cold-rolled 
shafting  is  made  from  the  same  material 
as  the  turned  shafting,  but  the  diameter 
is  reduced  by  rolling  cold  under  great 

pressure,  or  it  is  drawn  through  dies  which  reduce  its  diameter, 
when  it  is  known  as  cold-drawn  shafting. 

Shafting  Subjected  to  Torsion  Only.  If  a  shaft  carries 
no  gears  or  pulleys  and  transmits  power  it  will  be  subjected  to 
torsional  stress  only. 

Let    H  =  horse-power  to  be  transmitted, 
N= revolutions  per  minute  of  shaft, 
T  =  torque  in  inch-pounds, 
d  =  diameter  of  shaft  in  inches, 

P  =  tangential  force,  at  radius  R  inches,  to  be  trans- 
mitted.    (Fig.  6-1.) 

From  equation  (15)  chapter  I, 


FIG.  6-1. 


57 


58  ELEMENTS  OF  MACHINE  DESIGN 


but  for  a  circular  section  zP=-^d3,  then 


Solving  this  equation  for  d  we  obtain 


Generally  either  the  tangential  force  P  or  the  horse-power  H 
is  known.     The  value  of  T  in  terms  of  H  is  derived  as  follows: 

Horse    owcr-^or^  done  Per  mmute  in  foot-pounds 

ooOOO 

_PX  velocity  in  feet 
33000 


33000 
but 

PR=T; 
then 

2irTN 


12X33000' 


H  = 


"TV"";  ...........  (6) 

If  this  value  of  T  be  now  substituted  in  equation  (2)  the  result 
is 


Hollow  Shafts.  The  stress  produced  by  torsion  in  any  sec- 
tion of  a  shaft  is  not  uniformly  distributed  over  that  section,  but 
varies  directly  as  the  distance  from  the  center  of  the  shaft.  Then 
the  stress  is  a  maximum  at  the  outside  or  "  skin  "  of  the  shaft  and 
is  zero  at  the  center.  It  is  evident,  therefore,  that  the  material 
near  center  of  shaft  does  not  add  much  to  its  strength.  For  this 
reason  a  considerable  saving  in  weight  may  be  effected  by  the 


SHAFTS  AND  AXLES  59 

use  of  a  hollow  shaft,  and  large  shafts,  such  as  are  used  in  marine 
work,  are  frequently  made  hollow.  If  D  is  the  outside  diameter 
of  a  hollow  shaft  and  d  the  diameter  of  hole  the  section  modulus  is 


16V     D 
and  the  torque  which  such  a  shaft  can  transmit  is 


-Mn6r> 

If  ^  =  k  and  we  substitute  for  d  its  value  &D  we  obtain  from  (5) 


(6) 


EXAMPLE.  What  horse-power  can  a  hollow  steel  shaft  trans- 
mit when  rotating  at  120  R.P.M.?  The  outside  diameter  is  10  ins., 
inside  diameter  is  6  ins.  and  the  stress  is  not  to  exceed  10,000  Ibs. 
per  square  inch. 

From  (5) 

10000-  1296N 


10 

=  1,706,000  inch-pounds. 
From  (1) 

7W   =  1706000X120 

63025  63025 

=  3250. 

Shafts  Subjected  to  both  Bending  and  Twisting.  If  a  shaft 
carries  a  heavy  belt  wheel  or  gear  it  is  subjected  to  both  a  twist- 
ing moment,  due  to  power  which  is  transmitted,  and  to  a  bending 
moment,  due  to  weight  of  pulley  or  gear,  and  to  belt  pull  or  tooth 
pressure.  In  this  case  the  ideal  twisting  moment  is  obtained 
(equation  (17),  chapter  I). 


and 

(7) 


60 


ELEMENTS  OF  MACHINE  DESIGN 


EXAMPLE.  In  Fig.  6-2  the  shaft  transmits  40  H.P.  at  90 
R.P.M.  by  means  of  a  gear  60  ins.  diameter  placed  at  center. 
What  should  diameter  of  shaft  be  so  that  the  stress  will  not  exceed 
10,000  Ibs.  per  square  inch?  The  weight  of  gear  is  800  Ibs. 


63025  XH 

N 

63025X40 
90 


=  28,000  in.-lbs. 


The  tooth  pressure  is 


r_28000 

' 


FIG.  6-2. 


The  total  load  on  shaft  is  800+930=1730  Ibs.,  therefore  the 
bending  moment  is 

in.-lbs., 


and 


i  =  31200+  A/280002+312002 
=  73000  in.-lbs.; 


d=1.72 


3000 


10000 


=  3.34,  say  3f  in. 

Practical  Rules.  In  practice  it  is  frequently  impossible  to 
predetermine  the  distribution  of  loads  on  a  shaft  and  the  diameter 
is  calculated  from  the  horse-power  to  be  transmitted.  The  fol- 
lowing equation  may  be  used  for  steel  shafts: 


SHAFTS  AND  AXLES 


61 


Head  Shafts. 


Line  Shafts. 


Counter  Shafts. 


d 


3/100# 

"\    N    ' 


d  = 


N  ' 


40# 


N  ' 


(8) 

(9) 

(10) 


Proper  support  of  the  shafts  is  assumed  in  all  cases.  The  tor- 
sional  stresses  corresponding  to  the  above  equations  are  3200  Ibs. 
per  square  inch  for  head  shafts,  5350  Ibs.  per  square  inch  for  line 
shafts  and  8000  Ibs.  per  square  inch  for  counter  shafts. 

The  distance  between  center  of  bearings  of  line  shafts  depends 
on  the  diameter  of  the  shaft,  on  the  amount  of  transverse  stress 
due  to  belt  pull,  etc.,  and  to  some  extent  on  the  speed  of  shaft. 
For  average  conditions  the  following  table  gives  safe  values  of 
this  distance. 


Shaft  dia  

2" 

2*" 

3" 

3i" 

4" 

4*" 

5" 

Center  dist  

9'  6" 

10'  6" 

11'  0" 

12'  6" 

14'  0" 

16'  0" 

18'  0" 

Torsional  Rigidity  of  Shafts.  If  power  be  transmitted  to  a 
machine  through  a  long  line  of  shafting  the  angular  distortion  may 
have  a  disturbing  effect  on  the  proper  running  of  same  if  there  are 
sudden  variations  in  load.  It  is  therefore  usual  to  limit  this  dis- 
tortion, a  common  rule  being  that  the  angle  through  which  the 
shaft  may  be  twisted  is  not  to  exceed  J°  for  each  3  ft.  of  length. 
From  equations  (16)  Chapter  I: 


5  = 


57.2TL 
IPES   ' 


(ID 


Substituting  for  6  its  value  J°  and  for  IP,  ^d4,  for  E,  13,000,000 

(steel)  and  L,  36  ins.  we'obtain 

57.2TX36 


-d4X  13000000 


62  ELEMENTS  OF  MACHINE  DESIGN 

and 

^  =  .378^ (12) 

Axles.  An  axle  is  a  rotating  or  oscillating  bar  which  does  not 
transmit  any  torsional  moment  but  is  subjected  to  loads  which 
produce  bending  stresses.  Thus  in  Fig.  6-3 
the  axle  carries  a  bell  crank  and  the  resultant 
R  of  the  forces,  PI  and  P%,  produce  bending 
stresses  in  it.  The  axle  therefore  is  designed 
as  a  beam  of  circular  cross-section  carrying  a 
concentrated  load  R.  It  is  to  be  noted  that 
the  material  of  the  axle  is  subjected  to  a  con- 
tinually varying  stress  and  therefore  a  corre- 
FIG.  6-3.  spondingly  high  factor  of  safety  is  to  be  used. 

In  a  rotating  axle,  for  instance  car  axles, 
the  stress  will  vary  from  a  maximum  tension  to  a  maximum 
compression. 

PROBLEMS 

1.  What  horse-power  may  be  transmitted  by  a  shaft  2|  ins.  diameter 
at  140  R.P.M.  if  the  maximum  stress  is  10,000  Ibs.  per  square  inch?     Shaft 
is  subjected  to  torsion  only. 

2.  Find  the  diameter  of  a  shaft  which  will  transmit  a  torque  due  to  400 
Ibs.  acting  at  a  radius  of  18  in.     The  stress  not  to  exceed  6000  Ibs.  per  square 
inch. 

3.  A  shaft  3  in.  diameter  transmits  a  torque  of  16,000  in.-lbs.,  the  ulti- 
mate shear  resistance  of  the  shaft  material  being  45,000  Ibs.  per  square  inch. 
What  is  the  factor  of  safety? 

4.  What  horse-power  may  be  transmitted  by  a  hollow  shaft  at  90  R.P.M.? 
The  outside  diameter  is  15  ins.,  the  inside  diameter  is  .6  of  the  outside  diam- 
eter;  the  stress  to  be  8000  Ibs.  per  square  inch. 

5.  A  hollow  shaft  of  equal  strength  is  to  be  substitutec  for  a  solid  shait 
12  ins.  in  diameter.     The  inside  diameter  is  f  of  the  outside  diamete  .     Find 
diameter,  allowing  a  stress  of  9000  Ibs.  per  square  inch. 

6.  Compare  the  weight  of  a  hollow  shaft  18  ins.  outside  diameter  and 
10  ins.  inside  diameter  with  that  of  a  solid  shaft  of  equal  strength. 

7.  A  shaft  is  supported  on  two  bearings  8  ft.  between  centers.     It  carries 
a  pulley  3  ft.  from  one  bearing.     The  load  on  shaft  due  to  belt  pull  and  to 
weight  of  pulley  is  1100  Ibs.     The  shaft  transmits  transmits  25  H.P.  at 
120  R.P.M.     Find  diameter  of  shaft,  allowing  a  maximum  stress  of  8000 
Ibs.  per  square  in. 


SHAFTS  AND  AXLES 


63 


8.  What  horse-power  can  a  hollow  steel  shaft  transmit,  the  outside  diam- 
eter of  which  is  10  ins.,  inside  6  ins.  and  speed  120  R.P.M.?  The  safe  stress  is 
10,000  Ibs. 

9.  The  force  P  (1500  Ibs.)  at  lever  is  transmitted  to  lever  B.     Find  diam- 
eter of  shaft,  safe  stress  is  8000  Ibs.  per  square  inch.    Design  the  levers, 
assuming  them  to  be  of  rectangular  section,  depth  of  rectangle  three  times  the 
width  and  safe  stress  10,000  Ibs.  per  square  inch. 


CHAP.  VI.     Prob.  9. 


10.  Find  diameter  of  a  line  shaft  carrying  a  fair  proportion  of  pulleys.     It 
transmits  60  H.P.  at  150  R.P.M. 

11.  A  shaft  transmits  30  H.P.  at  120  R.P.M.     What  is  the  total  angle  of 
twist  if  the  shaft  is  60  ft.  in  length  and  2£  ins.  in  diameter? 

12.  A  hollow  steel  shaft  has  an  outside  diameter  of  9  ins.  and  inside 
diameter  of  6  ins.     Is  transmits  a  torque  of  300,000  in.-lbs.     Find  stress  in 
shaft. 

13.  A  shaft  is  supported  on  two  bearings  10  ft.  between  centers.     It 
carries  two  gears  of  60  and  36  ins.  diameter  respectively,  placed  3  ft.  from 
each  bearing.     The  shaft  transmits  80  H.P.  at  90  R.P.M.     Determine  the 
diameter  of  shaft,  allowing  a  stress  of  10,000  Ibs.  per  square  inch.     The 
weight  of  gears  is  900  Ibs.  and  500  Ibs.  each.     Assume  that  the  tooth  pressure 
acts  vertically  downward. 


CHAPTER  VII 
COUPLINGS  AND  CLUTCHES 

Couplings.  When  a  long  line  of  shafting  is  to  be  installed 
it  is  necessary  to  connect  the  ends  of  the  lengths  which  make  up 
this  line,  as  the  usual  length  of  a  piece  of  shafting  is  20  ft.  to 

1 


FIG.  7-1. 

25  ft.     The  machine  part  used  is  called  a  shaft  coupling.     There 
are  several  different  kinds  of  couplings  used  for  this  purpose. 

The  sleeve  or  muff  coupling  illustrated  in  Figs.  7-1  and  7-2  is 
simply  a  hollow  sleeve  keyed  to  the  shaft  ends,  the  first  being  a 


FIG.  7-2. 

solid,  while  the  latter  is  a  split  coupling.     The  length  L,  of  these 

couplings  may  be  made  3%d  and  the  outside  diameter  D  =  2d+l  in. 

For  large  shafts  flange  couplings  are  more  commonly  used. 

Fig.  7-3  illustrates  this  type.     The  following  proportions  may  be 

64 


COUPLINGS  AND  CLUTCHES 


65 


used:  Outside  diameter  D  =  3d+2  in.,  length  L  =  3d,  bolt  circle 
diameter  B  =  2\d.  The  bolts  are  in  shear  and  may  be  calculated 
on  the  assumption  that  they  are  capable  of  transmitting  the 


FIG.  7-3. 

same  torque,  T,  as  the  shaft,  and  that  the  load  is  carried  by  ^  of 
the  bolts. 


Let 
Then 

and 

if 

then 


i  =  bolt  diameter, 
n  =  number  of  bolts. 


2S'2; 


(1) 


(2) 


The  number  of  bolts  may  be  4  for  shafts  up  to  3 J  ins.  diameter, 
and  8  for  shafts  from  4  to  8  ins.  diameter. 

The  Sellers  coupling  (Fig.  7^4),  drives  by  means  of  friction 
between  the  conical  bushings  and  the  shaft.  These  bushings  are 
split  lengthwise  and  the  bolts  tighten  them  around  the  shaft, 
due  to  an  equal  taper  on  the  inside  of  sleeve.  The  length  L  may 


66 


ELEMENTS  OF   MACHINE  DESIGN 


be  3d+2  in.  and  the  outside  diameter  D  is  2d+2  in.     If  R  is  the 
mean  radius  of  the  bushings  the  tangential  force  at  this  circle  is 


P- 
- 


and  if  a  is  the  angle  of  the  bushings  the  axial  force  required  is 

Sin  a+n  cos  a 


where  /z  is  the  coefficient  of  friction.     The  load  on  each  bolt 
is  therefore  JQ.     The  angle  a  may  be  from  5°  to  10°. 


j 


FIG.  7-4. 


In  all  the  above  types  of  couplings  accurate  alignment  of 
shafts  is  essential.  These  couplings  should  be  placed  very 
close  to  a  bearing  and  as  far  from  pulleys  or  gears  as  possible, 
to  relieve  them  of  bending  stresses. 

The  universal  coupling  shown  in  Fig.  7-5  is  used  to  connect 
two  shafts  which  meet  at  an  angle  or  where  some  flexibility  is 
necessary  as  in  motor  cars.  It  should  be  remembered  that  with 
one  coupling  connecting  two  shafts  which  meet  at  an  angle  the 
transmission  of  motion  will  not  be  uniform  and  to  obtain  such 
motion  of  the  driven  shaft  two  couplings  must  be  used. 

A  flexible  coupling  is  advantageous  where  two  shafts  are 
to  be  coupled  and  accurate  alignment  cannot  be  maintained,  or 
if  a  prime  mover,  such  as  an  electric  motor,  is  coupled  directly 
to  a  shaft.  Figs.  7-6  and  7-7  show  two  types  of  this  coupling. 
In  the  first  the  connection  is  a  band  of  leather  or  cotton,  while 
the  latter  has  a  number  of  leather  links  connecting  the  two  halves 


COUPLINGS  AND  CLUTCHES 


67' 


of  coupling.     The  outside  diameter  D  may  be  made  5d,  the 
length  of  hubs  L  is  Ifd,  and  width  w  of  band  or  links  is  d. 

Clutches.  Couplings  which  may  be  readily  disengaged  are 
usually  termed  clutches.  These  are  used  not  only  to  connect  two 
shafts  but  also  to  couple  pulleys,  gears,  sprocket  wheels,  hoisting 
drums  and  similar  machine  parts  to  their  shafts.  The  variaticns 


FIG.  7-5. 

in  construction  are  very  numerous.  They  may,  however,  be 
divided  into  two  classes,  positive  clutches  and  friction  clutches, 
accordingly  as  the  motion  is  transmitted  by  positive  contact 
pressure  or  by  friction  between  surfaces.. 

Positive  Clutches..  The  jaw  clutch  shown  in  Figs.  7-8  and 
7-8a  is  the  commonest  type  of  positive  clutch.  One  part  of  this 
clutch  is  keyed  tightly  to  its  shaft  while  the  other  slides  on  a 
feather  key  and  therefore  can  be  moved  axially  until  its  teeth 
engage  with  those  of  the  other  half.  Fig.  7-8  shows  a  clutch 
capable  of  transmitting  rotation  in  both  directions,  while  that 


68 


ELEMENTS  OF  MACHINE  DESIGN 


shown  in  Fig.  7-8a  can  do  so  in  one  direction  only.  Such  clutches 
can  be  thrown  into  or  out  of  action  only  at  very  low  speeds. 
They  are  frequently  used  in  conveying  machinery.  The  sliding 


FIG.  7-6. 

member  for  large  clutches  should  have  two  feather  keys  placed 

180°  apart,  otherwise  it  is  difficult  to  move  same  along  the  shaft. 

Pin  clutches  are  of  the  positive  type  in  which  the  motion  is 


FIG.  7-7. 

transmitted  by  means  of  a  steel  pin.     They  are  largely  used  in 
punching  and  shearing  machinery. 

Friction  Clutches.      These  may  be  divided  into  cone,  dis<j, 
cylindrical  and  coil  clutches,  according  to  the  surfaces  in  con- 


COUPLINGS  AND  CLUTCHES 


69 


tact.  Figs.  7-9  to  7-12  show  examples  of  each  of  these  types. 
In  all  the  motion  is  transmitted  from  the  driving  part  to  the  driven 
by  means  of  friction  between  two  or  more  surfaces.  These  sur- 


FIG.  7-8. 

faces  are  pressed  together  with  sufficient  force  so  that  the  fric- 
tion between  them  will  overcome  the  resistance  to  motion  offered 
by  the  driven  member.  The  contact  surfaces  may  be  steel,  cast 
iron,  bronze,  brass,  wood,  cork,  leather  or  fiber. 


FIG.  7-8a. 


FIG.  7-9. 


Cone  Clutch.  This  clutch,  illustrated  in  Fig.  7-9,  consists  of 
a  cone  which  may  be  moved  along  the  shaft  until  it  comes  in  con- 
tact with  the  cup  into  which  it  fits. 

Let    Pt  =  the  tangential  force  to  be  transmitted, 

Pn  —  the  normal  pressure  between  friction  surfaces, 
Pa  =  axial  force  on  cone, 
R= mean  radius  of  cone  in  inches, 
H  =  coefficient  of  friction, 
H  —  horse-power  to  be  transmitted, 
N= revolutions  per  minute; 


70  ELEMENTS  OF  MACHINE  DESIGN 

then 

63025# 
1       RN    ' 

but  the  friction  between  contact  surfaces  must  be  at  least  equal 
to  the  tangential  force  to  be  transmitted;   then 


P0  =  Pn  sin  a 

=  P,^  .........     (3) 

This  would  give  the  axial  force  required  to  transmit  the  tan- 
gential force  Pt  or  to  engage  the  clutch  slowly,  neglecting  the  fric- 
tion which  must  be  overcome  in  moving  one  surface  over  the  other. 

The  angle  a  is  made  8°  to  12°  and  the  normal  pressure  per 
square  inch  is  10  to  40  Ibs.,  depending  on  the  material  of  the 
contact  surfaces  and  on  the  condition  of  service  for  which  clutch 
is  used.  High  speeds  and  soft  materials  require  the  lower  unit 
pressures. 

EXAMPLE.  Design  a  cone  clutch  to  couple  a  pulley  to  a  2-in. 
shaft.  This  pulley  transmits  10  H.P.  at  200  R.P.M. 

The  mean  radius  of  clutch  may  be  taken  at  4  to  5  times  the 
diameter  of  shaft,  say  8  ins.,  then 

63025X10 


8X200 


on*  IK 
=3951bs' 


For  wood  on  cast  iron  we  may  assume  /z  =  .2  and  a  may  be  12°, 
then 


Pn  =     =  1975  Ibs. 

M        .2 

Let  the  allowable  normal  pressure  be  20  Ibs.  per  square  inch,  then 
the  total  surface  of  contact  is 

£  =  -2Q-=100  sq.in.  nearly. 

Since  the  mean  radius  of  cone  is  8  ins.  the  width  of  cone  is 
.      S      100       . 


COUPLINGS  AND  CLUTCHES 


71 


The  total  axial  pressure  required  to  throw  the  clutch  into  engage- 
ment is 

Pt  sin  a. 


Pa  = 


=  395 


.208\  _ 

57" 


410  Ibs. 


In  the  disc  clutch  the  friction  surfaces  are  planes  perpen- 
dicular to  the  axis  of  the  shaft.  If  more  than  two  discs  are  used 
the  clutch  is  termed  a  multiple  disc  clutch. 


FIG.  7-10. 

In  Fig.  7-10  is  shown  a  disc  clutch  in  a  very  elementary  form. 
One  disc  is  rigidly  keyed  to  its  shaft  while  the  other  is  on  a  feather 
key  and  may  be  pressed  against  the  first  by  an  axial  force. 
Then  if 

p  =  normal  pressure  per  square  inch  between  discs, 
r  =  outside  radius  of  discs, 
Mf=  moment  of  the  friction  about  center  of  shaft  in  inch- 

pounds, 

T  =  torque  to  be  transmitted  by  clutch, 
Pa  =  total  axial  pressure. 
If  the  disc  is  assumed  to  extend  to  the  center  of  shaft,  then 


but  Mf  must  be  at  least  equal  to  the  torque  to  be  transmitted,  or 

(4) 


Where  much  power  is  to  be  transmitted  this  would  give  excessive 
axial  pressures;    to  avoid  this  the  multiple  disc  clutch  is  used. 

*  See  Appendix  A  for  development  of  this  equation. 


72 


ELEMENTS  OF  MACHINE  DESIGN 


Here  the  friction  surfaces  consist  of  two  sets  of  rings  as  shown  in 
Fig.  7-10a.  One  set  is  notched  at  outer  circumference,  these 
notches  fitting  over  an  equal  number  of  projections  on  outer  casing, 
thus  preventing  relative  rotation.  The  other  set  is  notched  at 


FIG.  7-10a. 

inner  circumference  and  a  corresponding  number  of  notches  on 
inside  drum  of  clutch  fit  into  these  notches,   the  discs  being 
assembled  alternately  one  of  each  set. 
Let  R  =  outside  radius  of  discs, 
r  =  inside  radius  of  discs, 
n  =  number  of  pairs  of  surfaces  in  contact. 


Then  approximately 
and 


R+r 
2     ; 


Pa  = 


(5) 
(6) 


Thus  by  increasing  the  number  of  pairs  of  friction  surfaces,  n,  we 
can  decrease  the  axial  thrust  Pa.  This  type  of  clutch  is  much 
used  in  automobiles  and  hoisting  machinery. 

In  the  cylindrical  clutch  (Fig.  7-11)  the  friction  surfaces 
are  cylinders,  or  portions  of  cylinders,  concentric  with  the  shaft. 
In  this  clutch  if  R  is  the  radius  of  friction  surface,  Pn  is  the  total 
normal  pressure  between  surfaces,  then 

T  ?  »PnR. 

The  methods  of  applying  the  normal  pressure  are  various,  gen- 
erally by  means  of  levers,  toggles,  or  screws.  This  type  of  clutch 
is  very  generally  used  in  shops  and  for  power  transmission. 


COUPLINGS  AND  CLUTCHES 


73 


The  coil  clutch  (Fig.  7-12)  consists  of  a  steel  coil  wound  on  a 
chilled  cast-iron  drum.  One  end  of  this  coil  is  fastened  to  a  sleeve 
or  hub  which  is  keyed  to  the  shaft  to  be  put  in  motion.  By 


FIG.  7-11. 

means  of  a  sliding  sleeve  and  a  lever  a  pull  is  exerted  on  the  free 
end  of  coil.  This  causes  it  to  tighten  around  the  drum,  each 
turn  increasing  the  friction  between  drum  and  coil.  In  Fig. 
12-13  is  shown  a  small  portion  of  the  coil.  The  force  required 


FIG.  7-12. 


FIG.  7-13. 


to  make  this  slip  to  the  left  is  the  force  tending  to  move  it  to 
the  right  plus  the  friction  between  this  portion  of  coil  and  the 
drum.  Thus  the  force  P  which  must  be  exerted  to  make  the 
band  or  coil  slip  over  drum  is  at  least 

P=Q+total  friction. 

Let  ju  be  the  coefficient  of  friction,  a  the  total  angle  in  radian 
measure,  subtended  by  coil,  and  e  the  base  of  the  Napierian 


74  ELEMENTS  OF  MACHINE  DESIGN 

logarithms  =  2.  7  18,  then  the  relation  between  P  and  Q  is  given 
by  the  equation 


In  this  equation  P  represents  the  pull  that  the  clutch  can  trans- 
mit if  a  pull  of  Q  Ibs.  is  exerted  at  loose  end  of  coil.  For  a  lubri- 
cated steel  band  on  a  cast-iron  drum  /z  may  be  taken  at  about 

lfcthen 

for  a  coil  of  1  turn  P  =  el/*Q  =  1.22  Q, 
for  a  coil  of  2  turn  P  =  e^5Q  =  1.5  Q, 
for  a  coil  of  4  turn  P  =  e4^Q  =  2.2  Q, 
for  a  coil  of  8  turn  P=e8/6Q=5  Q. 

PROBLEMS 

1.  Determine  the  stress  in  a  sleeve  coupling  for  a  2|-in.  shaft,  the  stress 
in  shaft  being  10,000  Ibs.  per  square  inch.     Take  dimensions  of  coupling 
according  to  empirical  equations  in  text. 

2.  Two  4-in.  shafts  are  coupled  by  means  of  a  flange  coupling.     Find 
diameter  of  bolts  if  there  are  six  of  them  and  the  stress  in  bolts  may  be  6000 
Ibs.  per  square  inch,  while  that  in  shaft  is  9000  Ibs.  per  square  inch. 

3.  Design  and  make  sketch  of  a  flange  coupling  for  a  hollow  shaft  12  ins. 
outside  and  6  ins.  internal  diameter. 

4.  What  horse-power  can  be  transmitted  by  a  cone  clutch  the  mean  radius 
of  which  is  8  ins.?     The  width  of  friction  surfaces  is  4  ins.  and  the  maximum 
normal  pressure  on  same  is  20  Ibs.  per  square  inch.     The  angle  of  cone  is 
10°  and  assume  /i  =  .12.     Shaft  runs  at  100  R.P.M. 

5.  Design  and  sketch  a  multiple  disc  clutch  to  transmit  20  H.P.  at  300 
R.P.M.     The  outside  diameter  not  to  exceed  14  ins.     The  discs  to  be  alter- 
nately steel  and  bronze.     State  all  necessary  assumptions. 

6.  In  a  cylindrical  clutch  the  diameter  of  the  friction  surface  is  18  ins. 
and  the  breadth  is  4  ins.     The  clutch  shoes  subtend  an  angle  of  180°  on  clutch 
rim.     The  pressure  per  square  inch  is  35  Ibs.     What  horse-power  can  be 
transmitted  at  100  R.P.M,  if  coefficient  of  friction  is  ,15? 


CHAPTER  VIII 
JOURNALS  AND  BEARINGS 

Journals.  That  part  of  a  rotating  spindle,  shaft,  or  axle, 
which  rests  in  a  support  is  called  a  journal,  while  the  support  itself 
is  the  bearing.  As  the  journal  is  part  of  the  shaft  the  calculation 
of  its  diameter  for  strength  is  a  part  of  the  design  of  the  shaft, 
and  this  was  considered  in  Chapter  VI.  It  is  rarely,  however, 
that  the  strength  of  the  journal  alone  is  the  important  consider- 
ation. The  fact  that  there  is  friction  between  a  journal  and  its 
bearings  and  that  therefore  heat  is  generated  is  largely  the  deter- 
mining factor  in  calculating  the  journal  dimensions.  The  bear- 
ing must  have  sufficient  area  to  dissipate  the  heat  generated  into 
the  surrounding  atmosphere,  otherwise  the  temperature  of  the 
bearing  would  soon  rise  to  a  point  where  the  lubricant  loses  its 
lubricating  qualities,  causing  destruction  of  both  journal  and 
bearing. 

Bearing  Pressure.  By  the  area  of  a  journal  is  meant  the  pro- 
jected area,  that  is,  the  cylindrical  area  of  the  contact  surfaces 
projected  on  a  plane  parallel  to  the  axis  of  the  shaft.  Thus,  if 
d  is  the  diameter  of  a  journal  and  I  is  its  length  both  in  inches 
the  projected  area  is  dl.  If  W  is  the  total  load  on  the  journal 
then  the  bearing  pressure  per  square  inch  of  projected  area  is 


W 


The  safe  value  of  p  depends  on  the  speed  of  rotation,  on  the  ma- 
terials used  for  journal  and  bearing,  on  the  quality  of  lubricant 
and  method  of  lubrication,  and  on  the  type  of  machine.  Table  1 
gives  values  of  p,  for  different  kinds  of  bearings,  as  commonly 
found  in  practice. 

75 


76  ELEMENTS  OF  MACHINE  DESIGN 

TABLE  1 


Kind  of  Bearing. 

p 

Lbs. 

Bearings  for  slow  speeds  and  intermittent  loads  

2000-4000 

Main  journals  of  steam  engines  

150-  300 

Crank-pins  of  high-speed  engines  

250-  600 

Crank-pins  of  low-speed  engines. 

700-1500 

Cross-head  pins  for  high-speed  engines  

800-1600 

Cross-head  pins  for  low-speed  engines 

1000-2000 

Motor  and  generator  bearings  

30-    80 

Line  shaft  bearings     

Heating  of  Journals.  The  amount  of  heat  generated  by  the 
friction  of  a  journal  depends  on  the  load,  the  rubbing  speed  and  the 
coefficient  of  friction  of  the  lubricant  and  surfaces.  As  the  rub- 
bing speed  increases  with  the  diameter  it  is  essential,  wherever 
possible,  that  this  be  kept  as  small  as  considerations  of  strength 
will  permit.  The  proper  length  of  a  journal  is  determined  by  its 
liability  to  heating. 

Let    d  =  diameter  of  journal  in  inches, 

1  =  length  of  journal  in  inches, 
N  =  re  volutions  per  minute, 
IL  =  coefficient  of  friction, 
W  =  total  load  on  journal; 

then  the  work  of  friction  is 


and  the  work  per  square  inch  of  projected  area  is 


therefore. 


The  amount  of  heat  generated  per  square  inch  of  projected  area 
in  British  thermal  units  is  the  work  of  friction  divided  by  778  or 


778~77SX!2l 


JOURNALS  AND  BEARINGS 


77 


With  well  constructed  and  properly  lubricated  bearings  a  value 
of  .02  may  be  assumed  for  /*;  this  gives' 

NW 


q=.  0000067 


I 


(6) 


The  value  of  q  varies  from  .2  to  1.0  or 

Z  =  0.000034#TF  to  0.0000067AW . 


The  ratio  of  length,  I,  of  journal  to  diameter,  d,  as  usually 
found  in  practice  is  given  in  Table  2.  In  calculating  the  length 
of  a  bearing  it  is  best  to  assume  this  ratio  from  the  table  and 
check  up  for  p  and  q  from  equations  1  and  6. 


TABLE  2 


Kind  of  Bearing. 

l 
d 

Transmission  bearings: 
Line  shaft,  countershaft,  etc  

3-5 

Heavy  pillow  blocks 

2i-3i 

Main  journals  of  steam  engines  

lf-2J 

Cross-head  pins  

1—2 

Crank-pins  ....                                                              ... 

3-u 

EXAMPLE.  A  shaft  carries  a  gear 
weighing  500  Ibs.  at  its  end  as  in  Fig.  8-1. 
The  pressure  on  gear  teeth  acts  vertically 
downward  and  is.  300  Ibs.  The  twisting 
moment  to  be  transmitted  is  6000  in. -Ibs. 
The  speed  is  200  R.P.M.  Design  a  suit- 
able bearing  (pillow  block).  The  bending 
moment  is  f 

M=  (300+500)10  =  8000  in.-lbs. 
The  ideal  twisting  moment  is 

Ti  =  8000+V^02+80002  =  18,000  in.-lbs.  ; 
then  the  diameter  of  shaft  and  journal  is 


J 



EE 

F 

id) 

<  —  10— 

IG.   8-1 

-•* 

J 

/1  soon 
d=l.72^  /i^™  =  2.16,  say  2J  ins. 


78  ELEMENTS  OF  MACHINE  DESIGN 

Assume  length  of  pillow  block  is  3d  or 


The  projected  area  of  journal  is 

s  =  2JX6f  =  15.19sq.in.; 
and  pressure  per  square  inch  of  projected  area  is 

P  =  TTTn  =  ^S,  which  is  a  safe  value. 

lo.iy 
The  heat  generated  per  square  inch  of  projected  area  is 

q  =  .  000067  ^=.16; 

i 

this  is  a  low  value  and  therefore  satisfactory. 

Bearings.     There   are   two   general   types   of  bearings,   viz.: 
journal  bearings  and  thrust  bearings.     A  journal  bearing  is  one 


FIG.  8-2. 

which  supports  a  load  acting  perpendicular  to  axis  of  shaft.  In 
thrust  bearings  the  load  acts  parallel  to  shaft  axis.  A  bearing 
sometimes  combines  these  functions. 

Steps  or  Brasses.  Except  for  very  low  speeds  or  very  light 
pressure,  bearings  are  always  lined  with  a  suitable  lining  called 
steps,  boxes,  or  brasses.  These  steps  are  usually  a  white  metal 
alloy  such  as  babbitt,  a  brass  or  a  bronze  alloy.  There  is  a  wide 
variation  of  bearing  metal  alloys  on  the  market  and  some  diversity 
of  opinion  as  to  their  respective  merits.  For  light  and  medium 
pressures  the  white  metal  alloys  are  preferred  and  are  commonly 


JOURNALS  AND  BEARINGS 


79 


used  for  transmission  bearings  and  the  main  journal  bearings 
steam  engines.  For  medium  and  heavy  pressures  and  also  for 
high  speeds  brass  or  bronze  linings  are  used,  the  latter  being  much 
the  better  of  the  two.  They  are  also  used  where  minimum  wear 
and  accurate  adjustment  are  important,  as  in  machine  tools. 
For  very  high  pressures,  particularly  in  thrust  bearings,  hardened 
steel  on  hardened  steel  is  the  best  combination.  The  thickness  of 
white  metal  linings  may  be 

Vd 

6  ' 
For  brass  or  bronze  steps 


t  = 


^  FIG.  8-3. 

Frequently  the  steps  are  of  cast  iron  or  brass  with  a  white  metal 
lining.  Figs.  8-5  and  8-6  shows  some  of  the  usual  types  of  steps. 
Lubrication.  To  prevent  destruction  of  journal  or  bearing 
it  is  essential  to  lubricate  them.  Perfect  lubrication  consists  in 
maintaining  an  unbroken  film  of  the  lubricant  between  the  journal 
and  its  bearing,  so  that  there  will  not  be  a  metal  to  metal  contact. 
The  lubricant  is  introduced  at  the  point  of  least  pressure,  which 
in  a  journal  bearing  is  usually  on  top.  The  simplest  method 
of  lubrication  is  to  cast  an  oil  well  in  cap  (Fig.  8-2)  and  feed  the 
lubricant  to  journal  by  means  of  a  wick.  A  better  way  is  to 
use  a  drip  oil  or  grease  cup  (Fig.  8-3) ;  this  permits  of  regulating 
the  quantity  of  lubricant  fed.  In  the  ring  oiling  bearing  (Fig. 


80 


ELEMENTS  OF   MACHINE  DESIGN 


8-8)  a  continuous  supply  of  oil  is  provided  by  means  of  one  or 
more  rings  (or  chains)  which  dip  into  a  reservoir  of  oil,  and  due  to 
their  rotation  with  shaft,  carry  the  oil  to  top  of  journal.  Prob- 
ably the  best  way  of  all  is  forced  lubrication.  In  this  system 


FIG.  8-4. 

a  raised  storage  oil  tank  or  a  small  pump  forces  the  requisite 
amount  of  lubricant  through  small  tubes  to  the  various  bearings 
of  the  machine.  This  system  is  largely  used  in  modern  steam 
and  gas  engine  practice  as  well  as  for  motor  cars. 


FIG.  8-5. 


FIG.  8-6. 


Oil  Grooves.  To  provide  proper  distribution  of  the  lubricant 
the  steps  are  provided  with  oil  grooves.  The  forms  of  these  are 
shown  in  Figs.  8-5  and  8-6.  Their  dimensions  range  from  a 
width  of  i*5  in.  and  depth  of  f  in.,  for  shafts  2  to  3  ins.  diameter, 
to  a  width  of  J  in.  and  depth  of  -f$  in.  for  shafts  of  18  to  20  ins. 
diameter.  Fig.  8-5  represents  the  brasses  of  a  two-part  bearing 
while  Fig.  8-6  shows  those  of  a  four-part  bearing. 

Adjustment  of  Bearings.  To  provide  an  adjustment  for  journal 
bearings  when  they  are  worn,  they  are  divided  into  two  or  more 
parts.  The  top  part  is  called  the  cap.  The  shaft  hanger  shown 
in  Fig.  8-6  has  the  hearing  divided  horizontally  into  two  parts. 


JOURNALS  AND   BEARINGS 


81 


The  post  bearing  (Fig.  8-7)  is  divided  on  a  45°  line.  Crank- 
shaft journal  bearings  are  often  divided  into  four  parts,  thus 
giving  both  up  and 
down,  and  a  side  wise 
adjustment. 

It  is  also  frequent- 
ly required  to  give 
the  bearing  a  slight 
motion  to  enable  it 
to  be  accurately  lined 
up  with  other  bear- 
ings. This  may  be 
done  by  supporting 
the  bearings  by  screws 
as  in  the  shaft  hanger 
(Fig.  8-8),  or  by  plac- 
ing it  on  a  sole  plate 
with  wedge  adjustment 
as  in  Fig.  8-4.  FlG>  8_7. 


FIG.  8-8. 


82 


ELEMENTS  OF   MACHINE  DESIGN 


Thrust  Bearings.     There  are  two  types  of  thrust  bearings, 
viz.:    pivot  bearings  and  collar  bearings.     In  the  pivot  bearing 


FIG.  8-9. 

the  end  of  the  shaft  is  the  bearing  surface,  while  in  the  collar 
bearing  one  or  more  collars  are  formed  on  shaft  and  the  thrust 


FIG.  8-10. 


is  taken  by  them.     Fig.  8-9  illustrates  the  simplest  type  of  pivot 
bearing  in  which  the  end  of  the  shaft  rests  on  a  hardened  steel 


JOURNALS  AND  BEARINGS 


83 


washer.  Fig.  8-10  shows  a  better  form  of  bearing  in  which  a 
separate  end  piece  of  hardened  steel  is  let  into  the  shaft  and  a 
number  of  washers  are  used;  these  are  alternately  steel  and  brass. 


FIG.  8-11. 

Collar  bearings  are  used  chiefly  oa  horizontal  shafts  which 
have  to  sustain  a  large  axial  thrust,  as  the  propeller  shafts  of 


FIG.  8-12. 


steamboats.  Fig.  8-11  shows  a  collar  bearing  for  a  vertical 
shaft  and  Fig.  8-12  one  for  a  horizontal  propeller  shaft  for  a  small 
steamer. 


84  ELEMENTS  OP  MACHINE  DESIGN 

Friction  Work  in  Pivot  and  Collar  Bearings.     It    may    be 

assumed  that  the  pressure  is  distributed  uniformly  over  the  sur- 
face.    Then  for  pivot  bearings: 

Let    p  =  pressure  in  pounds  per  square  inch, 
R  =  radius  of  bearing  surface  in  inches, 
W  —  total  load  in  pounds, 
N  =  revolutions  per  minute, 
Mf=  moment  of  friction  in  inch-pounds, 
n  —  coefficient  of  friction  ; 

then 

W 

P=^ 
and 

Mf=lnRW.     (See  Friction  disc  clutch.) 

The  work  of  friction  is 

A  =  27rN%^~  =  .35»RNWftAbs.       ...     (7) 

For  collar  bearings  or  pivot  bearings  in  which  the  contact  sur- 
face is  a  ring, 

Let  R  =  outside  radius  in  inches, 

r  =  inside  radius  in  inches, 
n  =  numbers  of  collars; 
then 

W 
P 


From  these  equations  it  is  evident  that  the  work  lost  in  fric- 
tion increases  with  the  radius  R  and  it  is  therefore  advantageous 
to  keep  the  radius  as  small  as  possible.  In  collar  bearings  the 
diameter  of  collars  may  be  from  1.3d  to  1.6d,  where  d  is  the 
diameter  of  shaft.  Table  3  gives  safe  values  of  p  for  pivot  and 
collar  bearings.  These  values  must  be  taken  as  very  approxi- 
mate as  information  on  this  subject  is  widely  varying. 


JOURNALS  AND  BEARINGS  85 

TABLE  3 

Lbs. 
Very  low  speeds: 

Column  cranes,  turn-tables,  etc 1500-2000 

Turbine  foot-step  hearings : 

Very  high  grade  construction 700-1200 

Collar  bearings: 

Propel^    shafts 60-90 

EXAMPLE.  The  thrust  bearing  of  a  steam  yacht  is  to  be 
designed.  The  engine  is  500  I.H.P.  and  the  propeller  9  ft.  pitch, 
the  shaft  being  5J  ins.  diameter.  The  engine  makes  220  R.P.M. 

If  we  assume  a  slip  of  20  per  cent  the  boat  will  be  driven  7.2 
ft.  for  each  revolution  of  propeller  or  1580  ft.  per  minute.  But 
the  pressure  on  thrust  bearing  times  the  speed  in  feet  per  minute 
is  the  work  done  per  minute  in  foot-pounds.  In  designing  thrust 
bearing  this  is  equated  to  the  I.H.P.  of  the  engine.  Then 

1580  XTF  =  500X33,000, 
W=  10,450  Ibs. 

The  outside  diameter  of  collars  may  be  8  ins.,  and  p  we  will 
assume  to  be  60  Ibs.     Then  each  collar  can  support  a  thrust 

W'=^  (82- 5^)60=  1600  Ibs.  nearly; 

and  therefore  the  number  of  collars  required  is 

10450 

w==T^7sr  — 6.55,  say  7  collars. 
IbUO 

The  work  lost  in  friction  is 

Z3_9l3 
A  =  .35M220X  10450-2     Ii2; 

A* 9— 

and  if  bearing  is  lubricated  in  oil  bath  p  may  be  assumed  at 
.01,  then 

'  A=41300ft.-lbs. 


86  ELEMENTS  OF  MACHINE  DESIGN 

Ball  Bearings.  The  heavy  friction  losses  occasioned  by  ordi- 
nary journal  bearings  has  led  to  the  development  of  ball-and-roller 
bearings.  In  these  types  rolling  friction  is  substituted  for  sliding 
friction.  Ball  bearings  were  first  applied  extensively  in  the  bicycle. 
Until  the  publication  of  Prof.  Stribeck's  research  work  it  was 
commonly  held  that  ball  bearings  were  suitable  only  for  very 
light  loads.  They  are  now  used  in  great  variety  of  machines 
under  all  conditions  of  load  and  speed. 

Radial  ball  bearings  are  those  supporting  loads  at  right  angles 
to  the  axis  of  the  shaft.  Fig.  8-13  shows  a  radial  bearing  consist- 


FIG.  8-13. 

ing  of  a  single  row  of  balls  placed  between  an  inner  and  an  outer 
ring  called  races.  The  races  may  be  straight  or  grooved,  the 
radius  of  the  groove  being  about  two-thirds  the  ball  diameter. 
In  order  to  assemble  this  bearing  it  is  necessary  to  cut  a  notch 
at  right  angles  to  the  groove.  In  these  bearings  the  balls  are  in 
contact  with  each  other  and  since  the  points  of  contact  move  in 
opposite  direction  there  is  considerable  friction  and  wear.  To 
obviate  this  the  balls  are  separated  by  a  light  metal  cage  as  shown 
in  Fig.  8-14,  this  being  now  the  almost  universal  custom  for 
medium  and  heavy-load  ball  bearings. 

Allowable  Loads.  Radial  Bearings.  The  total  load  which 
a  radial  ball  bearing  may  safely  carry  depends  on  the  number  of 
balls  and  their  diameter. 


JOURNALS  AND   BEARINGS 

Let  P  =  total  load, 

p  =  load  on  one  ball, 
n  =  number  of  balls  in  bearing, 
d  =  diameter  of  balls  in  J  in., 
fc  =  a  constant. 


87 


FIG.  8-14. 

The  load  may  be  considered  as  being  carried  by  one-fifth  the 
number  of  balls  in  bearing  or 


The  safe  load  for  one  ball  is 


(10) 


the  value  of  k  may  be  taken  from  10  to  15  for  first-class  material 
and  grooved  races,  and  about  one-half  of  these  figures  for  straight 
races.  The  above  equations  do  not  consider  speed  of  rotation, 
and  this  would  not  affect  the  carrying  capacity  if  it  were  absolutely 
uniform  and  the  load  steady.  Such  conditions,  however,  are  rarely 
met  with  in  practice,  and  therefore  with  increasing  speeds  and 
varying  loads  the  capacity  should  be  decreased. 

In  angular  bearings  the  constraining  surfaces  are  at  an  angle 
to  the  axis  of  the  shaft.     Thus  Figs.  8-15,  8-16,  8-17,  show  re- 


88 


ELEMENTS  OF   MACHINE  DESIGN 


spectively  two-,  three-  and  four-point  contact  bearings.     In  order 
to  insure  true  rolling  of  the  balls  it  is  essential  that  the  lines  through 


FIG.  8-15. 


FIG.  8-16. 


the  points  of  contact  intersect  in  a  point  on  the  axis  of  the  shaft 
or  that  they  be  parallel  to  this  axis. 


I 


§ 


L-l 


FIG.  8-17. 


FIG.  8-18. 


The  importance  of  high-grade  material  and  workmanship 
can  hardly  be  overestimated.  The  material  for  both  balls  and 
races  is  a  special  grade  of  cast  steel.  This  must  be  of  uniform 


JOURNALS  AND  BEARINGS 


hardness  and  structure.  The  balls  are  highly  polished  so  that  no 
grinding  scratches  can  be  detected.  In  the  best  bearings  their 
diameters  do  not  vary  by  more  than  .0001  in. 

Experiments  have  shown  that  the  coefficient  of  friction  is 
almost  independent  of  the  speed  of  rotation.  Its  mean  value 
may  be  taken  at  .0015  for  a  normally  loaded  bearing. 

Ball  Thrust  Bearings.  These  are  used  for  all  loads  and  speeds. 
For  very  low  speeds,  as  in  the  steering  gear  of  motor  cars,  turn 


1,500 


1,000 


500 


1,000 


2,000 
Load  .in  Ibs. 

FIG.  8-19. 


3,000 


tables,  cranes,  etc.,  the  races  are  filled  with  balls,  for  higher  speeds 
cages  are  used  as  shown  in  Fig.  8-18.  In  thrust  bearings  the  total 
load  is  of  course 

P  =  np (11) 

This  assumes  a  uniform  distribution  of  the  load  on  all  the  balls. 
To  insure  this  it  is  usual  to  use  a  spherical  seat  for  the  thrust 
washer  supported  by  the  fixed  frame.  In  thrust  bearings  the 
safe  load  per  ball  decreases  rapidly  with  increasing  speeds.  Thus 
in  the  equation  p  =  kd2  the  value  of  k  is  no  longer  constant  but 


90 


ELEMENTS  OF   MACHINE  DESIGN 


depends  on  the  speed  of  shaft.  Just  what  the  variation  in  k 
should  be  is  not  definitely  known  and  we  can  only  depend  upon  the 
accumulated  experience  of  the  manufacturers  of  such  bearings. 


FIG.  8-20. 

Fig.  8-19  shows  safe  loads  for  speeds  from  10  to  1500  revolu- 
tions per  minute  for  a  thrust  bearing  containing  eighteen  J-in. 
balls,  as  recommended  by  a  well-known  manufacturer.  This 
shows  values  of  k  from  4.6  to  33.4. 


FIG.  8-21. 

Cleanliness  is  essential  to  the  success  of  all  forms  of  ball 
bearings.  They  must  be  thoroughly  protected  from  grit  and  dirt. 
If  a  shaft  does  not  pass  through  the  bearing  the  outer  end  is  closed 


JOURNALS  AND  BEARINGS 


91 


by  a  dust-proof  cover.  Where  the  shaft  passes  through  the 
bearing  frame  should  be  bored  out  -^  to  ^  in.  larger  than  shaft, 
and  a  groove  turned  in  this  bore  as  shown  in  Fig.  8-20.  A 
small  hole  is  drilled  on  the  inside  to  communicate  with  the  oil 
space;  the  edges  of  this  groove  must  be  sharp.  Sometimes  a 
felt  washer  is  used  to  make  a  dirt-tight  closure.  The  bearings 


FIG.  8-22. 

should  be  lubricated  with  a  good  quality  of  oil,  free  from  acid 
which  will  not  attack  the  highly  polished  surfaces  of  balls  or  races. 
Roller  Bearings.  In  the  simplest  type  of  roller  bearing  the 
space  between  shaft  and  housing  is  filled  with  rollers.  There  is 
in  this  type  of  bearing,  however,  a  tendency  for  the  rollers  to  get 
out  of  alignment,  which  causes  rapid  destruction  of  the  bearing. 
For  this  reason  the  rollers  are  more  usually  carried  in  a  suitable 


FIG.  8-23. 

cage  as  shown  in  Fig.  8-21.  in  the  Hyatt  roller  bearing  the 
rollers  are  made  of  a  strip  of  steel  rolled  into  a  cylinder.  This 
gives  sufficient  flexibility  to  the  roller  to  insure  contact  along  the 
entire  length  of  same,  resulting  in  a  uniform  distribution  of  the  load 
on  it  as  well  as  the  surface  on  which  it  rolls.  Roller  bearings 
have  also  been  used  as  thrust  bearings.  The  rollers  may  be  coni- 
ical  (Fig.  8-22)  or  if  cylindrical  divided  into  a  number  of  short 
sections  (Fig.  8-23). 


92  ELEMENTS  OF   MACHINE  DESIGN 

Design.     The  safe  load  on  a  roller  depends  on  its  diameter 
and  length. 

Let    d  =  roller  diameter  in  inches, 
I  =  roller  length  in  inches, 
n  =  number  of  rollers  in  bearing, 
v  =  velocity  of  roller  in  feet  per  minute, 
P  =  total  safe  load; 
then 

P  =  knld2. 

The  value  of  k  depends  on  the  velocity  v;  the  following  table  gives 
average  practice: 

t>(feet  per  minute)  =0  to  20     100       200       300      400       500 
k  =     200         50         22         17        13         10 

PROBLEMS 

1.  The  main  journals  of  a  steam  engine  are  6  ins.  diameter  by  10  ins.  in 
length.     The  load  supported  on  each  bearing  is  14,000  Ibs.     The  speed  of 
engine  is  120  R.P.M.     Find  horse-power  lost  in  friction  at  each  bearing  if 
.««.04. 

2.  A  shaft,  supported  on  two  bearings  7  ft.  between  centers,  carries  a 
gear  located  2  ft.  from  one  of  the  bearings.     The  load  on  shaft  due  to  weight 
of  gear  plus  tooth  pressure  is  22,000  Ibs.     Design  the  journals  so  that  the 
pressure  per  square  inch  of  projected  area  is  180  Ibs.  and    assume    ratio 
l:d=2l 

3.  In  a  steamboat  the  total  thrust  on  the  propeller  thrust    bearing  is 
50,000  Ibs.     The  engine  develops  3000  H.P.  at  180  R.P.M.     The  propeller 
shaft  is  hollow  with  the  hole  f  of  the  outside  diameter.     Find  diameter  of 
shaft,  allowing  a  stress  of  8000  Ibs.  per  square  inch.     Determine  the  number 
of  thrust  collars  required  if  their  outside  diameter  is  1|  shaft  diameter  and  the 
bearing  pressure  is  75  Ibs.  per  square  inch. 

4.  Design  the  foot-step  bearing  for  a  column  crane.     The  maximum  load 
on  crass  is  20  tons  and  the  weight  of  crane  is  8  tons. 


CHAPTER  IX 
BELTS  AND  PULLEYS 

Belts  are  made  of  various  materials,  such  as  leather,  cotton, 
hemp  and  rubber.  Leather  is  by  far  the  commonest  and  best 
material,  oak-tanned  ox-hide  being  used  for  the  highest  grade 
belts.  The  strongest  part  of  the  hide  is  a  strip  12  to  16  ins. 
wide  and  about  4|  ft.  in  length,  along  the  back  of  the  animal. 
This  strip  has  the  most  uniform  elastic  qualities,  its  thickness 
is  from  .20  to  .25  in.  For  narrow  and  inferior  belts  the  side  of 
the  hide  is  used ;  the  thickness  of  this  part  of  the  hide  being  from 
.25  to  .35  in.  In  manufacturing  a  belt  these  strips  are  beveled 
off  at  the  ends,  glued,  sewn  or  riveted  together  until  the  desired 
length  is  obtained.  Belts  may  be  single,  double  or  triple,  accord- 
ing to  there  being  one,  two  or  three  thicknesses  of  leather. 

Cotton,  hemp  and  rubber  belts  are  used  chiefly  in  exposed  or 
damp  locations.  They  are  cheaper  than  leather  belts,  but  do 
not  last  as  long.  Cotton  belts  are  either  woven  endless  or  made 
by  sewing  together  4  to  10  thicknesses  (plies)  of  canvas  or  duck. 
Rubber  belts  are  made  by  cementing  several  plies  of  canvas  with 
a  rubber  composition. 

Belt  Fasteners.  There  are  many  different  methods  of  join- 
ing the  ends  of  leather  belts.  The  best  method,  and  the  one 
usually  adopted  for  large  belts  and  important  drives,  is  to  scarf 
the  ends  and  cement  them  together.  This  makes  the  joint  nearly 
as  strong  as  the  rest  of  the  belt.  In  small  belts  the  joint  is  gen- 
erally made  by  lacing  the  ends  together  with  rawhide  or  wire. 
This  joint  is  only  from  one-half  to  two-thirds  as  strong  as  the 
belt.  It  readily  permits  of  taking  up  the  stretch  in  new  belts  and 
as  this  amounts  to  about  6  per  cent  of  its  length,  it  has  to  be  done 
more  or  less  frequently  in  long  belts  in  order  to  maintain  the 
proper  driving  tension.  There  are  a  number  of  patented  metallic 
belt  fasteners  on  the  market,  some  of  which  are  shown  in  the  belt 
joints  (Fig.  9-15). 

93 


94  ELEMENTS  OF  MACHINE  DESIGN 

Belt  Transmission.  Fig.  9-1  represents  two  pulleys  connected 
by  a  belt.  It  is  evident  that  in  order  to  produce  rotation  of  the 
follower  the  moment  of  the  friction  between  belt  and  pulley  about 
center  B  must  be  at  least  equal  to  the  moment  of  the  resistance  to 
be  overcome.  This  friction  is  due  to  the  initial  tension  put  on  the 
belt  when  it  is  plated  on  the  pulleys,  or,  in  order  words,  the 


Driver 
FIG.  9-1. 

tension  in  belt  when  at  rest.  Let  T%  be  this  tension.  When  the 
driver  starts  to  move  it  pulls  on  the  driving  or  tight  side  of  belt, 
increasing  the  tension  there  and  decreasing  that  of  the  opposite 
or  slack  side  until  the  difference  between  these  tensions  is 
sufficient  to  overcome  the  resistance  to  motion  offered  by  the 
follower.  Then  if 

TI  =  tension  in  tight  side, 
T%  =  tension  in  slack  side, 
P  =  driving  force  exerted  at  rim  of  pulley; 
we  have 


(1) 


P=Ti-T2.       .     ......     (2) 

The  difference  of  TI  and  T2  is  equal  to  the  friction  of  the  belt 
on  the  pulley  rim.  The  relation  between  them  being  given  by 
the  equation 


where  e  is  the  base  of  the  natural  logarithms  (e  —  2.718),  6  is 
the  angle  of  contact  of  belt  and  pulley  rim  expressed  in  radian 
measure  (180°=7r  radians)  and  /^  is  the  coefficient  of  friction. 

*  See  appendix  B  for  derivation. 


BELTS  AND  PULLEYS 


95 


Since  T\  is  the  maximum  tension  in  belt  it  will  determine  the 
necessary  cross-sectional  area  and  therefore  the  width  required; 
and  if 

A  =  sectional  area  of  belt  in  square  inches, 
st  —  safe  tensile  stress  in  pounds  per  square  inch ; 
then 

Ti  =  Astt (4) 

or  if  the  safe  tension  per  inch  of  width  be  designated  by  k  and 
width  of  belt  by  6,  then 

(5) 


J  +  dT 


FIG.  9-2. 

The  following  table  gives  values  of  &**  for  various  values  of 
M  and  6: 


PORTION  OF  CIRCUMFERENCE  IN  CONTACT. 


M 

.2 

.3 

.4 

.45 

.50 

.55 

.6 

.7 

.25 

.37 

1.60 

1.87 

2.03 

2.19 

2.37 

2.57 

3.00 

.28 

.42 

1.69 

2.02 

2.21 

2.41 

2.63 

2.81 

3.43 

.33 

.51 

1.86 

2.29 

2.54 

2.82 

3.13 

3.47 

4.27 

.38 

.61 

2.05 

2.60 

2.93 

3.30 

3.72 

4.19 

5.32 

.40 

.65 

2.13 

2.73 

3.10 

3.51 

3.98 

4.52 

5.81 

.50 

.87 

2.57 

3.51 

4.11 

4.81 

5.63 

6.59 

9.00 

The  angle  6  should  be  the  smaller  angle  subtended  by  a  belt 
on   a   pair   of   pulleys.     For   an   open   belt   this   will   be   6°  = 


180°  —  2 


—  j 


where  R  and  r  are  the  radii  of  the  two 


pulleys  and  C  is  the  distance  between  centers. 


96 


ELEMENTS  OF  MACHINE  DESIGN 


In  the  above  equations  the  effect  of  centrifugal  force  has 
been  neglected.  For  belt  speeds  above  2500  ft.  per  minute 
this  force  should  be  taken  into  account.  Its  effect  is  to  increase 
the  tension  in  the  belt  without  increasing  its  driving  power  and 
therefore  the  safe  value  of  T\  should  be  reduced  by  an  amount 
equal  to  the  centrifugal  force. 

Let  w  =  weight  of  1  cu.in.  of  leather, 
R  =  radius  of  pulley  in  inches, 
g  =  acceleration  of  gravity  in  feet  per  second, 
c  =  centrifugal  force  of  1  in.  length  of  belt, 
v  =  velocity  of  belt  in  feet  per  second; 

then 

_wAv2 

~H~R' 

12 

The  sum  of  all  these  forces  (Fig.  9-3)  in  a  direction  parallel  to 
the  belt  is 


FIG.  9-3. 

and  therefore  the  tension  due  to  centrifugal  force  produced  in 
belt  per  square  inch  of  cross-sectional  area  is 

C      I2wv2 


The  average  weight  of  leather  may  be  taken  at  w  =  .035  lb., 
<7=32.2; 
then 


BELTS  AND   PULLEYS  97 

and  therefore 


.......     (6) 

from  which  required  width  of  belt  may  be  calculated. 

The  value  of  the  coefficient  of  friction,  ju,  varies  between  rather 
wide  limits,  depending  on  the  material  of  the  belt  and  of  the  pul- 
leys, on  the  amount  of  slip  of  the  belt,  and  to  some  extent  on  its 
speed.  An  average  value  for  a  leather  belt  on  a  cast-iron  pulley 
is  .33.  For  pulleys  made  of  pulp  or  wood  it  is  somewhat  less, 
except  when  new,  and  for  paper  pulleys  considerably  more. 

The  breaking  strength  of  good  leather  belts  varies  from 
3500  to  4500  Ibs.  A  safe  stress  of  300  Ibs.  per  square  inch  for 
belts  in  which  the  ends  are  cemented  together  is  good  practice, 
although  the  smaller  this  stress  the  longer  the  life  of  the  belt. 

EXAMPLE.  A  15-H.P.  motor  drives  a  line  shaft.  The  motor 
makes  600  R.P.M.  and  has  a  12-in.  pulley  on  armature  shaft. 
Find  width  of  double  belt  required  if  .4  of  pulley  circumference 
is  in  contact  with  belt. 

The  belt  velocity  is 

TX12X600     lonnr 
v  =  —  —  r»  --  =  1890  feet  per  minute; 

therefore  the  effect  of  centrifugal  force  may  be  neglected.     The 
tangential  force  at  rim  of  pulley  is 

15X33000 
1890 

If  n  be  assumed  at  .33  then  from  the  table 

1  =  ^=2.29, 
then 


.'.     Ti=4551bs. 

This  is  the  maximum  tension  which  occurs  in  belt.  The  thickness 
of  a  double  belt  is  about  f  ".  Allowing  a  tension  of  300  Ibs.  per 
square  inch  the  cross-sectional  area  required  is 


98 


ELEMENTS  OF  MACHINE  DESIGN 


and  therefore  the  width  is 

1  52 

&=-^=4.06-ins.,  or  say  4-in.  belt. 
.0/0 

Practical  Methods.  It  is  not  always  possible  to  predeter- 
mine the  exact  conditions  under  which  a  belt  is  to  run,  and  various 
practical  rules  are  in  use  for  determining  the  width  of  a  belt  to 
transmit  a  given  horse-power.  In  general  it  will  be  found  that 

7^  =  2.5  gives  satisfactory  results;  then 

12, 

If  b  is  the  width  of  belt  in  inches  and  k  is  the  allowable  tension 
in  belt  per  inch  of  width  then 

6=T  =  £         -    -     (7) 

but 


P_ 

*  ~~ 


and 


H.P.X  33000 


55000  XH.P. 
vk 


(8) 


The  thickness,  t,  of  belts,  the  effective  pull,  P,  per  inch  of 
width  and  the  corresponding  maximum  tension,  k,  per  inch  of 
width,  on  tight  side  of  belt,  as  found  in  American  practice,  is 

given  in  table  below.     In  obtaining  k  it  was  assumed  that  -^  =  2.5. 

12 


t 

P 

k 

Single  belt 

A-5& 

25-40 

42-  67 

Double  belt  

&-H 

35-60 

58-100 

Triple  belt. 

1  6_  ?  2 

55-80 

92-142 

A  rough  rule  of  thumb  is  to  allow  65  sq.ft.  of  belt  to  pass  over 
pulley  for  each  horse-power  transmitted,  or,  if  A  is  belt  area, 
passing  over  pulley,  in  square  feet  per  minute,  we  have 


TTP    =A—]?Lt 

65    780' 


BELTS  AND  PULLEYS  99 

and 

.       780XH.P.  800XH.P.  f  . 

b=  or  sav .      .     .         .     (y) 

v  v 

For  a  double  belt  40  sq.ft.  may  be  allowed  per  horse-power  or, 

up  -A=  &L. 
n         40    480' 

and 

,     480  H.P.  500XH.P.  f    . 

b=—          -  or  say  -  — (10; 

v  v 

General  Rules  for  Installation.  Belts  may  be  used  to  connect 
two  pulleys  wherever  a  small  variation  of  angular  velocity  ratio 
is  of  no  consequence.  For  best 
results  there  are  certain  limitations 
as  to  center  distance  between  shafts. 
If  the  .distance  is  very  short  a  tight 
belt  is  required,  which  shortens  its 
life  and  increases  the  wear  on  shaft 
bearings.  One  rule  is  to  make  the 
minimum  distance  between  centers 
twice  the  diameter  of  the  larger 
pulley.  Belts  are  rarely  used  for 

center  distances  of  more  than  60  ft.,  as  it  is  cheaper  to  install 
rope  drives  for  longer  distances. 

Horizontal  drives  are  better  than  vertical.  This  applies 
especially  to  main  drives,  as  the  machines  in  a  shop  must  of  neces- 
sity have  nearly  vertical  drives.  The  driving  side  should  be 
on  bottom,  as  this  gives  the  largest  arc  of  contact.  The  necessary 
initial  tension  may  be  obtained  by  any  one  of  three  methods: 
first  by  making  the  belt  short  enough  so  that  it  has  to  be  stretched 
over  the  pulleys,  thus  producing  the  tension.  Second,  by  the 
use  of  tightening  pulleys  or  idlers  (Fig.  9-7).  These  should  be 
applied  on  the  slack  side  of  belt.  Third,  in  the  case  of  large  belts 
and  long  center  distances  the  weight  of  the  belt  itself  is  sufficient 
to  produce  the  required  tension. 

When  two  shafts  which  are  not  parallel  are  to  be  connected 
by  a  belt  it  may  be  necessary  to  use  guide  pulleys.  In  any  case 
the  following  rule  must  be  observed,  viz.:  the  center  line  of  the 
advancing  side  of  the  belt  must  be  in  a  plane  which  is  perpendicular 


100  ELEMENTS  OF  MACHINE  DESIGN 

to  the  shaft  toward  which  it  is  advancing.     Fig.  9-5  shows  what 
is  termed  a  quarter  twist  drive  connecting  two  shafts  at  right 


FIG.  9-5.  FIG.  9-6. 

angles.  Here  rotation  in  one  direction  only  is  possible.  Fig. 
9-6  shows  a  similar  drive  with  the  use  of  two  guide  pulleys  per- 
mitting rotation  in  either  direction. 


FIG.  9-7. 


Creep  in  Belts.     When  two  pulleys  are  connected  by  a  belt 
the  theoretical  angular  velocity  ratio  of  driver  to  that  of  driven 

is  T^ri  wnere  D  ig  diameter  of  driver,  d  diameter  of  driven,  and 


t  is  the  thickness  of  belt.     Actually  the  speed  of  driven  pulley 
will  be  2  to  3  per  cent  less  than  the  theoretical  speed.     This  loss 


BELTS  AND  PULLEYS  101 

is  due  to  creep  of  belt.  Creeping  results  from  the  elasticity  of 
belt.  If  there  be  no  slipping  of  the  belt  on  pulley,  its  rim  will 
have  the  same  velocity  as  that  of  belt  where  contact  begins. 
If  c  is  the  coefficient  of  extension;  that  is,  the  amount  a  belt  1  ft. 
long  stretches  when  subjected  to  a  stress  of  1  Ib.  per  square  inch, 
then  each  original  foot  of  belt  as  it  runs  off  at  h  (Fig.  9^)  will 

have  a  length  1+^jC  ft.     At  g  this  length  will  be  l+^|c  ft. 
A.  A. 

Therefore  the  difference  in  length  of  belt  running  on  at  g  and 

rn    rp 

off  at  h  is  -       — -  c.     This  represents  the  proportionate  loss  in 
A. 

speed  and  reducing  to  simplest  terms 

Tl-T2_P_ 
A      ~A~P' 

that  is,  p  represents  the  driving  force  of  belt  per  square  inch  of 
sectional  area.  Then  the  proportional  loss  of  speed  is 

L  =  pcv. 

This  loss  of  speed  must  not  be  confused  with  slipping  of  belt 
due  to  excessive  load. 

PULLEYS 

Material  and  Construction.  The  materials  used  for  pulleys 
are  cast  iron,  steel,  wrought  iron,  wood  pulp,  and  paper,  cast 
iron  being  by  far  the  commonest  material  used.  They  may  be 
solid,  that  is,  in  one  piece,  or  split  pulleys  made  in  two  halves  and 
bolted  together.  Pulleys  running  at  high  speed  should  be  care- 
fully balanced  to  prevent  excessive  vibration.  The  rims  may  be 
flat  (Fig.  9-8),  crowned  (Fig.  9-9)  or  flanged  (Fig.  9-10).  The 
flat  rim  is  used  where  a  belt  has  to  be  shifted.  The  purpose  of 
crowning  is  to  make  the  belt  run  centrally  on  pulley.  Flanged 
pulleys  are  used  where  the  belt  may  fall  off  due  to  excessive  slip- 
ping or  on  vertical  shafts. 

Proportions  of  Rims,  Anns  and  Hubs.  The  width,  w,  of 
pulley  rim  (Fig.  9-8)  should  be  greater  than  that  of  belt  and  may 
be  made 

in, 


102  ELEMENTS  OF  MACHINE  DESIGN 

The  amount  of  crowning  is 


c  = 


w 


10' 


The  thickness  of  rim  at  edge  may  be 

VD  .   1 
<=-20-+l6in" 

where  D  is  diameter  of  pulley  in  inches.     The  diameter  of  hub 

is  given  by 

in. 


T 


FIG.  9-9. 


FIG.  9-10. 


FIG.  9-8. 

where  d  is  the  diameter  of  shaft. 

The  arms  may  be  calculated  on  the  assumption  that  they  are 
cantilevers  subjected  to  a  load  equal  to  the  effective  belt  pull. 
Let    P  =  pull  of  belt  at  rim, 

R  =  pulley  radius  in  inches, 
n  =  number  of  arms. 

Then  we  may  assume  that  actually  one-half  of  the  arms  carry 
the  load  P  and  therefore  the  bending  moment  on  each  arm  is 

M=PRJ*PR 

\n 

M=zs: 
and 

2PR 


n 


2  = 


ns 


BELTS  AND  PULLEYS  103 

With  the  usual  elliptic  section  of  arm  (Fig.  9-8)  in  which  major 
axis,  a,  is  twice  the  minor  axis,  6,  the  section  modulus  is 


and  substituting  this  value  in  equation  (11)  we  obtain 


128PR 

irns 

? 

(12) 

ns 

This  gives  the  dimensions  at  center  of  pulley,  the  arm  being 
tapered  from  \  to  f  in.  per  foot.  However,  the  dimensions  at 
rim  should  not  be  less  than  two-thirds  of  those  at  center.  Very 
broad  pulleys  are  provided  with  two  sets  of  arms. 

EXAMPLE.  Determine  dimensions  of  arms  of  a  cast-iron  pulley 
48  in.  diameter,  transmitting  30  H.P.  at  150  R.P.M.  The  pulley 
has  six  arms. 


The  safe  stress  may  be  taken  at  from  1800  to  3000,  using  2000 
we  obtain  from  equation  (11) 

31 


Tight  and  loose  pulleys  are  shown  in  Figs.  9-11.  These  are 
used  where  a  machine  is  to  be  started  and  stopped  frequently, 
although  in  a  great  many  cases  a  friction  clutch  is  to  be  preferred 
for  this  purpoSST  The  hub  of  the  loose  pulley  is  made  quite  long 
to  reduce  wear.  The  pulley  may  run  directly  on  shaft  or  a  cast- 
iron,  steel,  brass,  bronze  or  babbitted  bushing  may  be  used,  as  in 
Fig.  9-12.  Here  the  sleeve  is  held  on  shaft  by  a  set  screw,  it  is 
hollowed  out  to  act  as  an  oil  reservoir.  Sometimes  the  diameter 
of  the  loose  pulley  is  somewhat  less  than  that  of  the  tight  pulley, 
the  purpose  being  to  relieve  the  tension  in  belt. 

Split  pulleys  are  easy  to  put  in  position  on  shaft.  Fig.  9-13 
shows  a  cast-iron  split  pulley.  The  weight  of  steel  pulleys  is  only 
about  one-third  that  of  cast  iron.  This  may  be  of  considerable 


104 


ELEMENTS  OF  MACHINE  DESIGN 


advantage  for  line  shafts  with  many  pulleys,  since  there  is  a  corre- 
sponding reduction  of  bearing  friction  and  wear. 

Cone  Pulleys  are  used  when  a  number  of  speeds  are  required, 
as  in  nearly  all  machine  tools.  It  is  advantageous  to  have  the 
speeds  in  geometrical  progression. 


FIG.  9-13. 


Let 


then 


nx  =  speeds  of  driven  shaft, 
r  =  ratio  of  geometric  progression, 
x  =  number  of  steps  on  cone  pulley, 

n2  =  rni,  n3  =  rn2  =  r2n,  etc.; 


BELTS  AND  PULLEYS 


105 


and 


/nA_L 

=  (-?*-: 
V»i/ 


Since  the  belt  is  shifted  from  one  step  of  the  cone  to  the  other 
it  ig  of  course  necessary  that  their  diameters  be  such  as  to  require 
the  same  length  of  belt  on  each  step.  Fig.  9-14  gives  Burmester's 


FIG.  9-14. 


graphical  solution  of  this  problem.  The  distance  between  centers 
of  shafts  and  the  radii  of  the  first  step  of  each  of  the  cone  pulleys 
must  be  known. 

Draw  the  line  A  C  at  an  angle  of  45°  to  the  horizontal  line  AB 
and  length  equal  to  d,  the  distance  between  shafts.  Draw  CD 
equal  to  \d  and  perpendicular  to  AC.  With  A  as  a  center  and 
radius  AD  strike  an  arc.  Locate  a  point  E  on  this  arc  so  that 
the  vertical  distance  between  E  and  a  point  F  on  AC  is  equal  to 
R— r,  the  difference  of  the  radii  of  the  first  steps  of  pulleys.  Ex- 
tend this  line  to  G  so  that  FG  is  equal  to  r.  Through  G  draw  the 
horizontal  line  GO,  then  EG  =  R  and  OG=FG  =  r.  Draw  OE 

FC1     f? 

and  let  angle  EOG=B,  then  tan  0  =  -— =  — =  the  velocity  ratio 

m.  The  radii  Ri,  and  n,  for  any  other  velocity  ratio  mi,  may  be 
found  by  drawing  OEi,  so  that  tan  0i=mi,  then  EG\  =  Ri  and 


106  ELEMENTS  OF  MACHINE  DESIGN 


PROBLEMS 

1.  A  pulley  54  ins.  diameter  transmits  60  H.P.  at  175  R.P.M.     The  arc 
of  contact  is  160°.     Find  width  of  A-in.  belt  which  would  be  required  if 
coefficient  of  friction  is  .35  and  a  safe  stress  of  300  Ibs.  per  square  inch  be 
permitted. 

2.  An    engine    develops    50  H.P.  at  200  R.P.M.      The    belt  wheel    is 
60  ins.  diameter  and  drives  through  an  intermediate  jack  shaft,  a  dynamo 
running  at  1200  R.P.M.  and  having  an  18-in.  pulley  on    armature    shaft. 

ip 
Find  width  of  both  belts  assuming  —=2.5  and  allowing  a  tension  of  90  Ibs. 

^2 
per  inch  of  width.  Sketch  the  arrangement  of  pulleys  giving  their  diameters. 

3.  What  horse-power  can  be  transmitted  by  a  belt  4  ins.  in  width  and 
£  in.  thickness  when  running  over  a  26-in.  pulley  making  250  R.P.M.,  the 
coefficient  of  friction  is  .38  and  the  maximum  stress  not  to  exceed  300  Ibs. 
per  square  inch.     Arc  of  contact  is  160°. 

4.  Design  and  make  sketch  of  a  pulley  40  ins.  in  diameter  to  transmit 
30  H.P.  at  125  R.P.M.     The  pulley  bore  is  3  ins. 

5.  An  engine  band  wheel  is  84  ins.  in  diameter.     It  transmits  100  H.P. 
at  125  R.P.M.     It  has  8  arms  of  the  usual  elliptic  section.     Determine  dimen- 
sions of  arms  allowing  a  maximum  bending  stress  of  3000  Ibs.  per  square 
inch. 


CHAPTER  X 
FRICTION  WHEELS 

Friction  Wheels  are  applicable  for  light  and  medium  powers 
to  machinery  which  is  to  be  frequently  stopped  and  started,  also 
where  a  wide  range  of  speeds  is  desired  as  in  the  feed  mechanism 


FIG.  10-1. 


FIG.  10-2. 


1 


of  various  machine  tools. .  For  power  transmission  a  fairly 
high  rotative  speed  is  essential.  The  friction  surfaces  may  be 
metal,  wood,  paper,  fiber,  leather,  or  rubber.  The  driver  should 
be  of  the  softer  material,  the  driven  being  a 
metal  wheel,  usually  cast  iron.  Spur,  bevel 
and  disc  friction  wheels  are  commonly  em- 
ployed as  shown  in  Figs.  10-1  to  10-3. 

As  the  driving  capacity  of  these  wheels 
depends  on  the  friction  of  the  contact  surfaces, 
it  is  essential  to  keep  these  clean  and  free 
from  grease  or  oil.  They  should  be  rigidly 
supported  by  adjacent  bearings  to  maintain 
an  even  contact  pressure  across  the  entire 
face  of  the  wheels. 

Transmission   of   Power.     The   coefficient 
of  friction  and  the  contact  pressure  determine 
the  amount  of  power  that  can  be  transmitted. 
The    conditions    vary  so    widely    that    laboratory   experiments 
.on  the  coefficient  of  friction  do  not  always  give  a  safe  value  for 

107 


FIG.  10-3. 


108 


ELEMENTS  OF   MACHINE  DESIGN 


their  design.     In  the  table  below  /*  is  the  coefficient  of  friction 
and  p  is  the  safe  contact  pressure  per  inch  of  width  of  face. 


Material. 

n 

p 

Cast  iron  on  cast  iron.        .            

10-  15 

300-500 

Wood  on  cast  iron 

20-  50 

150-200 

Paper  on  cast  iron 

15-  22 

150-250 

Leather  on  cast  iron    

.20-.  30 

200-300 

Rubber  on  cast  iron 

20 

Fiber  on  cast  iron  

.20-.  30 

150-250 

Let  w  =  width  of  cylindrical  friction  wheel  in  inches, 
/i  =  coefficient  of  friction, 
D  =  diameter  of  wheel  in  feet, 
N  =  revolutions  per  minute, 
p  =  safe  contact  pressure  per  inch  of  width, 
P  =  tangential  force  at  rim  of  wheel, 
V  =  velocity  of  contact  surfaces  in  feet  per  minute; 
then 

P  =  nwp 

and  the  horse-power  transmitted  is 

TT_ 
" 


33000       33000 
=  .  000095  ftwpND  ......     (1) 

It  is  more  usual  that  the  width  of  a  wheel  to  transmit  a  given 
horse-power  is  required,  solving  equation  (1)  for  this,  we  obtain 

H 


Bevel  Frictions  are  suitable  only  for  very  light  power  trans- 
mission, as  it  is  difficult  to  maintain  an  even  contact  pressure. 
The  forces  Qi  and  $2  (Fig.  10-4)  which  must  be  exerted  along  the 
shafts  to  engage  the  wheels  and  to  obtain  a  driving  force  P  are 

sin  CE+    cos  a. 


p 
"8* 


/0v 
W 


(4) 


FRICTION  WHEELS 


109 


and  the  horse-power  transmitted  is 

#  =  . 000095PM); (5) 

where  D  is  the  mean  diameter  of  either  bevel  wheel  and  N  the 
revolutions  per  minute  of  the  same  wheel. 

Grooved  Frictions  are   capable  of  transmitting  more  power 
than  cylindrical  and  are  therefore  used  where  considerable  power 


— 


FIG.  10-4. 


FIG.  10-5. 


is  required,  as  in  hoisting  machines.     If  a  is  one-half  of  the 
groove  angle  (Fig.  10-5)  then  using  the  same  notation  as  before 

,sin  OL+H.  cos  a 


Q=P' 

M 

The  usual  values  of  a  are  12°  to  20°. 
iron  wheels  with  /i  =  .12.^ 

Q  =  3.2P. 


(6) 

For  a  =  15°  and  two  cast- 


The  efficiency  of  this  class  of  friction  wheels  may  be  taken  at 
from  85  to  90  per  cent.  As  there  can  be  rolling  motion  at  one 
point  only  on  each  line  of  contact  there  is  considerable  wear. 
To  reduce  this  as  far  as  possible  the  depth  of  groove  should  be 
small,  from  f  to  f  in. 

Methods  of  Engaging  Friction  Wheels.  Fig.  10-6  shows 
the  method  of  engaging  two  spur  frictions.  The  bearing  sleeve 
A  is  bored  eccentrically  and  may  be  rotated  through  a  small  angle 


110 


ELEMENTS  OF   MACHINE  DESIGN 


by  means  of  the  attached  lever.  In  this  way  the  center  C  of  the 
shaft  is  shifted  and  the  two  wheels  may  be  pressed  together. 
For  bevel  wheels  the  bearing  sleeve,  A,  has  a  long  pitch  thread 
cut  on  its  outside  as  shown  in  Fig.  10-7.  This  sleeve  may  b'e  turned 


FIG.  10-6. 


FIG.  10-7. 


by  means  of  the  lever  L,  thus  moving  it  in  the  direction  of  the 
shaft  axis.  The  axial  motion  of  sleeve  is  transmitted  directly  to 
bevel  wheel  or  to  shaft  by  means  of  set  collars. 

PROBLEMS 

1.  Two  shafts  24  ins.  between  centers  are  connected  by  a  pair  of  friction 
wheels.     The  driver  makes  240  R.P.M.,  the  follower  making  120  R.P.M. 
Find  width  of  friction  if  15  H.P.  is  to  be  transmitted.     Let  the  pressure  be 
150  Ibs.  per  inch  of  width  and  the  coefficient  of  friction  =  .2. 

2.  A  small  printing  press  is  driven  by  a  3-H.P.  motor  running  at  1200 
R.P.M.     On  the  motor  shaft  is  a  rubber  friction  wheel  5  ins.  diameter.     Find 
width  of  friction  wheel  if  the  pressure  is  100  Ibs.  per  inch. 

3.  A  pair  of  grooved  frictions  is  used  to  connect  two  shafts  33  ins.  between 
centers.     The  driver  makes  240  R.P.M.  and  the  follower  200  R.P.M.     Find 
pressure  on  bearings  if  12  H.P.  is  to  be  transmitted  and  ju  =  -12,  angle  a  =  15°. 

4.  In  problem  3  if  each  wheel  has  six  grooves  £  in.  deep,  what  is  the  con- 
tact pressure  per  inch  of  contact  line? 

5.  Find  horse-power  that  can  be  transmitted  by  a  grooved  friction  wheel 
18  ins.  diameter  when  running  at  250  R.P.M.     The  wheels  are  pressed  to- 
gether with  a  force  of  1000  Ibs.  and  the  coefficient  of  friction  is  .2.    The 
angle  of  grove  (a)  is  15°. 

6.  A  certain  friction  hoisting  machine  has  two  driving  wheels  1\  ins. 
diameter.     The  two  followers,  which  are  44  ins.  diameter,  are  keyed  to  the 
drum  shaft.     The  faces  are  6  ins.     Determine  the  load  which  can  be  hoisted 
if  the  drum  diameter  is  23  ins.     Assume  pressure  per  inch  of  face  is  250  Ibs. 
and  coefficient  of  friction  is  .2.     What  horse-power  is  required  if  the  drivers 
make  225  R.P.M.? 


CHAPTER  XI 


TOOTHED  GEARS 

Spur  Gears.  Toothed  gearing  is  used  to  transmit  motion 
between  shafts  which  are  comparatively  close  together,  or  where 
a  definite  velocity  ratio  between  the  shafts  must  be  accurately 
maintained,  as  in  the  screw-cutting  mechanism  of  a  lathe.  The 
imaginary  surfaces  upon  which  the  teeth  are  formed  are  called 
pitch  surfaces.  The  gears  connecting  parallel  shafts  have  cyl- 
indrical pitch  surfaces  and  are  called  spur  gears. 

The  object  of  the  teeth  is  to  obtain  a  positive  transmission 
of  motion.  The  tooth  profiles  must  be  such  that  the  motion  trans- 
mitted by  them  will  give  the  same  veloc- 
ity ratio  as  would  be  obtained  by  rolling 
the  pitch  surfaces  upon  each  other.  It  is 
assumed  that  the  reader  is  familiar  with 
the  methods  of  drawing  gear  teeth  of 
either  the  involute  or  the  cycloidal  sys- 
tems. Gear  teeth  may  be  cast  or  cut  by 
properly  formed  cutters.  Since  cast  gears 
are  more  or  less  inaccurate  they  are  not 
suitable  for  high  speeds  on  account  of 
excessive  noise  and  vibration. 

Strength  of  Gear  Teeth.  In  the  following  method  of  calcula- 
tion a  number  of  assumptions  are  made  which  are  more  or  less 
approximately  correct.  In  Fig.  11-1  let 

W  =  load  acting  on  tooth  in  pounds, 
p  =  circular  pitch  of  gear, 
b  =  breadth  (face)  of  gear  in  inches, 
h  =  height  of  tooth  =  .7p, 
t  =  thickness  of  tooth  =  .5p. 

The  tooth  acts  as  a  cantilever  carrying  the  load  W  at  its  end, 
then  the  bending  moment  is 

.7pW. 
Ill 


FIG.  11-1. 


112  ELEMENTS  OF   MACHINE  DESIGN 

The  section  at  which  this  bending  moment  acts  is  a  rectangle  of 
height  t  and  breadth  b. 


and 

16.8TP 


It  is  usual  to  assume  the  breadth,  6,  of  the  teeth  in  terms  of 
the  pitch  ;  that  is,  b  =  np  where  n  is  a  constant  whose  value  depends 
on  the  speed  of  the  gears.  Then 

_  16.8TF 


IW 
/.     p  =  4.lJ—  .       .......     (2) 

\ns 

For  very  low  speeds  as  in  hand-operated  machines,  n  =  lj  to  2J, 
for  medium  and  high  speeds  n  =  3  to  6.  In  very  high-speed 
machinery  such  as  steam  turbines  where  herringbone  gears 
are  used  the  value  of  n  may  be  from  20  to  30,  or  more. 

It  frequently  happens  that  the  load  W  is  not  known,  but 
the  twisting  moment  T  and  the  number  of  teeth,  N,  are  known. 
Then  if  R  is  the  pitch  radius  of  the  gear  in  inches 

T=WR; 
but 


== 

R     Np' 
If  this  value  of  W  be  substituted  in  equation  (2)  we  obtain 


nNps 

31 


(3) 


TOOTHED  GEARS 


113 


If  the  horse-power,  H,  to  be  transmitted  and  the  speed,  w, 
in  revolutions  per  minute,  are  known,  then  since 


H 

:.     W 


33000  X12 
HX33000X12 


2irR< 


Nv 
substituting  for  R  its  value  •=*-, 

Zir 


W 


ffX33000X12 


This  value  of  W  may  now  be  substituted  in  equation  (2)  and  we 
obtain 

'33000  X 12  XH 


pnsN( 


(4) 


In  deducing  the  above  equations  it  has  been  assumed  that 
the  entire  force  comes  on  one  tooth  and  that  it  acts  at  right  angles 
to  the  radial  plane  through  center  of  tooth.  A  third  assumption, 
that  the  thickness  of  tooth  at  the  root  is  equal  to  one-half  of  the 


FIG.  11-2. 


pitch  is  true  only  for  one  wheel  in  an  interchangeable  set.  The 
tooth  thickness  being  less  in  pinions  having  few  teeth  and  greater 
in  gears  having  many  teeth.  This  is  shown  in  Figs.  11-2,  where 
a  is  the  tooth  of  a  96-tooth  gear,  and  b  that  of  a  10-tooth  pinion. 


114 


ELEMENTS  OF  MACHINE  DESIGN 


EXAMPLE.  Design  the  gears  for  a  hand-operated  hoist  to 
lift  a  load  of  1500  Ibs.,  assuming  drum  to  be  8  ins.  in  diameter 
(Fig.  11-3). 


301bs. 


FIG.  11-3. 

If  the  crank  be  12  ins.  long  and  the  force  exerted  on  crank 
handle  is  30  Ibs.  the  gear  train  ratio,  neglecting  friction,  is 

=  4X  1500  ^16.6 
g     30X12        1   ' 

Using  four  gears  this  ratio  may  be  divided  into  two  factors  or 
_16.6_4    4.15 

nr~ix  i 

and  the  number  of  teeth  may  now  be  assumed  according  to  this 
ratio 

5v^=??v^ 
C*A  -  20A12* 

Equation  (3)  will  be  applicable  to  this  case.     Assume  n  =  2  and 
s  =  6000,  then  since  for  gear  D  the  twisting  moment  is 

T  =  4X  1500  =  6000  in.-lbs., 


p=4. 


6000 


2X83X6000 
?86  =  say  Jin. 


TOOTHED  GEARS  115 


For  gear  B  the  twisting  moment  is 


m,       6000 

T'  =   -r  =  1445  m.-lbs.  ; 


and 


=  1445 


2X48X6000 
=  .64  in.  =  say  f  in. 

The  Lewis  Equation.  This  equation  for  the  strength  of  gear 
teeth  deduced  by  W.  Lewis  takes  into  consideration  both  the  form 
of  the  tooth  and  the  obliquity,  of  the  line  of  pressure.  It  is  widely 
used  by  American  designers.  Using  the  same  notation  as  pre- 
viously this  equation,  for  the  15°  involute,  and  for  the  cycloidal 
system  in  which  the  12-tooth  pinion  has  radial  flanks  is 


(5) 

and  substituting  for  6  its  value  np  we  obtain 

P=    /-     —  (6) 


If  the  twisting  moment  T  to  be  transmitted  is  known  we  have 

> 

=  ~Np' 
and  substituting  this  value  of  W  in  (6), 


JT 
-557T-      ..-     (7) 
nsAM.124-- 


The  values  of  the  safe  stress  s  for  any  material  depends  to  a 
great  extent  on  the  speed  of  the  gears  and  on  the  steadiness  of  the 
load.  The  following  table  gives  safe  values  for  gears  that  are 
well  supported  in  rigid  bearings.  For  excessive  shock,  as  in  rock- 
crushing  machinery  or  rolling-mill  work,  these  values  may  be 
reduced. 


116 


ELEMENTS  OF   MACHINE  DESIGN 


Velocity  of  Pitch  Line, 
Feet  per  Minute. 

100  or 
Less. 

200 

300 

600 

1000 

1500 

2000 

s  for  cast  iron  

6,000 

4,500 

3,600 

3000 

2500 

2000 

1800 

s  for  stssl 

15,000 

12,000 

10,000 

9000 

8000 

6000 

5000 

For  the  cycloidal  system  in  which  the  15-tooth  pinion  has    radial 
flanks  the  Lewis  equation  is 


(8) 


Bevel  gears  are  used  to  connect  two  shafts  which  lie  in  the 
same  plane  but  are  not  parallel  to  each  other.  Such  gears  have 
conical  pitch  surfaces.  Fig.  11^  shows  a  pair  of  bevel  gears  to 


FIG.  11-4. 

connect  two  shafts  at  right  angles.  The  pitch  of  these  gears  may 
be  calculated  in  the  same  way  as  that  of  spur  gears.  The  value 
thus  found  is  to  be  taken  as  the  mean  pitch  half  way  between  the 
large  and  the  small  end  of  the  teeth  at  radius  R.  The  actual 

pitch  at  large  end  is  -^  times  the  calculated  pitch. 
H 

Spiral  gears  are  used  to  connect  two  shafts  that  are  neither 
parallel  nor  in  the  same  plane.  Fig.  11-5  shows  a  pair  of  such 
gears.  The  tooth  elements  here  are  helices  and  therefore  the 
gears  are  in  point  contact,  theoretically.  For  this  reason  the 


TOOTHED  GEARS 


117 


strength  of  the  tooth  is  rarely  a  factor  in  their  design  as  the 
contact  pressure  limits  the  safe  load. 

Worm  Gears.  ,The  worm  and  worm  wheel  may  be  used  where 
a  large  speed  reduction  is  required;  the  shafts  usually  are  at  right 
angles.  The  pitch  of  worm  wheel  is  calculated  as  that  of  a  spur 
gear.  However  since  in  worm  gears  there  are  always  a  number 
of  pairs  of  teeth  in  contact,  it  will  be  safe  to  assume  that  the 
load  on  teeth  to  be  used  in  the  calculation  for  pitch  is  \  of 
the  total  tangential  force  acting  at  pitch  line  of  worm  wheel. 


FIG.  11-5. 

Fig.  11-6  shows  a  section  through  a  pair  of  these  gears.  Since 
the  teeth  slide  over  each  other  with  a  velocity  equal  to  that  of 
the  pitch  line  of  the  worm,  wear  is  an  important  consideration 
in  their  design.  To  limit  the  wear  it  is  therefore  necessary  to 
reduce  the  contact  pressure  with  increasing  speed. 

The  efficiency  of  worm  gearing  depends  on  the  angle  of  the 
worm  thread,  that  is  on  the  angle  of  the  pitch  helix  of  worm. 
Let   E  —  efficiency, 

a  =  angle  of  worm  thread, 

/*  =  coefficient  of  friction, 

d= pitch  diameter  of  worm, 

0= tan"1  /* 


118  ELEMENTS  OF  MACHINE  DESIGN 

then  if  the  friction  of  bearings  be  neglected 

tan  a 


E  = 


tan 


(9) 


From  this  equation  it  is  evident  that  E  =  0  when  a  =  0  and  also 
when    a  =  90°— /?.     The    efficiency    being    a    maximum    when 

a  =  — ~— :-.     For  cast-iron  worm  wheel  and  a  steel  worm  running 


FIG. 


in  oil  and  good  workmanship,  /*  may  be  taken  at  .05.  This 
gives  for  maximum  efficiency  a  =43°  30'  approximately.  So 
large  an  angle  cannot  be  obtained  in  practice  and  fortunately 

a 


the   efficiency  is  very   satisfactory,   if  a.  is  from  15°  to  20° 

value  which  may  be  obtained  by  proper  design.     Since  tan  a  =  —, 

ird 

it  is  evident  that  d  should  be  made  as  small  as  possible  in  order  to 
obtain  the  maximum  efficiency  for  a  given  pitch. 


TOOTHED  GEARS 
THEORETICAL  EFFICIENCY 


llfl 


a 

5° 

10° 

15° 

20° 

25° 

30° 

.02 

81.3 

89.5 

92.6 

94.1 

95.0 

95.5 

.04 

68.4 

80.9 

86.1 

88.8 

90.4 

91.4 

.06 

59.0 

73.8 

80.4 

84.0 

86.1 

87.5 

.08 

51.9 

67.8 

75.4 

79.6 

82.2 

83.8 

Gear  Rims  and  Arms.  The  thickness,  t,  of  gear  rims  may  be 
made  .5p+J  in.  for  gears  of  small  pitch  and  Ap  for  gears  of  large 
pitch.  Fig.  11-7  shows  the  common  sections  of  gear  arms.  They 
may  be  calculated  in  the  same  way  as  the  pulley  arms  in  Chap- 


FIG.  11-8. 

ter  IX,  except  that  the  assumption  is  made  that  only  one-third 
of  the  arms  carry  the  load.  This  changes  equation  (11)  of  that 
chapter  to 

3PR 


120  ELEMENTS  OF  MACHINE  DESIGN 

from  which  the  proportions  of  the  arms  at  the  hub  may  be  cal- 
culated. The  diameter  of  the  hub  may  be  D  =  1.75d+i  in.  where 
d  is  the  shaft  diameter. 

Split  Gears.  Large  and  heavy  gears  are  frequently  cast 
in  halves  and  these  are  bolted  together  or  shrink  links  are  used. 
Fig.  1 1-8  shows  a  portion  of  such  a  gear'  which  is  parted  along  the 
center  of  an  arm.  The  bolts  should  be  calculated  so  that  those  at 
the  rim  alone  will  be  capable  of  sustaining  the  entire  force  acting 
at  the  teeth.  In  very  heavy  gears  both  shrink  links  and  bolts 
are  used. 

PROBLEMS 

1.  A  gear  having  a  pitch  diameter  of  6  ins.  is  to  transmit  5  horse-power 
at  120  R.P.M.     Determine  its  pitch  if  a  stress  of  4000  Ibs.  per  square  inch 
be  permitted  and  face  of  gear  is  3£  times  the  pitch. 

2.  Two  shafts  running  at  90  and  126  R.P.M.  and  transmitting  70  H.P. 
are  connected  by  spur  gears.     The  distance  between  centers  is  to  be  as  near 
60  ins.  as  possible.     Find  pitch  of  gears,  assuming  face  to  be  three  times  the 
pitch  and  allowing  a  stress  of  4500  Ibs.  per  square  inch. 

^j  3.  A  gear  has  40  teeth  of  2  diametral  pitch  and  4  ins.  face.  It  transmits 
a  twisting  moment  of  15,000  in.-lbs.  Determine  stress  in  teeth. 

4.  A  punch  press  running  at  40  R.P.M.  is  driven  by  a  10-H.P.  motor  run- 
ning at  1200  R.P.M.,  a  double-gear  reduction  being  used.     Design  the  gear 
train,  state  all  necessary  assumptions.     Make  a  sketch  of  mechanism. 
\f    5.  Design  the  arms  for  gears  in  problem  2,  assume  each  gear  has  six  arms 
and  that  the  stress  is  3500  Ibs.  per  square  inch. 

6.  The  gate  of  a  sluice  valve  weighing  10,000  Ibs.  is  raised  by  means  of  a 
rack  and  pinion.     Design  a  train  of  gears  so  that  gate  may  be  lifted  by 
two  men  working  on  a  14-in.  crank  handle  and  exerting  a  pressure  of  40  Ibs. 
each. 

7.  The  drum  of  a  hoist  is  16  ins.  in  diameter.     The  gear  on  same  shaft  with 
drum  is  36  ins.  in  diameter  and  meshes  with  a  pinion  6  ins.  in  diameter.     On 
same  shaft  with  this  pinion  is  a  gear  24  ins.  in  diameter  which  meshes  with 
another  pinion  6  ins.  in    diameter. ,  The  capacity  of  hoist  is  3000  Ibs.     De- 
termine pitch  and  number  of  teeth  on  each  gear.     Assume  stress  at  5000  Ibs. 
and  face  of  each  gear  2^  times  the  pitch. 

8.  An  18-tooth  cast-iron  pinion  is  to  transmit  15  H.P.  at  200  R.P.M. 
Determine  the  pitch  required  by  means  of  the  Lewis  equation.     Assume 
n  =  3i 

\J  9.  In  a  worm  gear  of  1-in.  pitch,  the  worm  is  double  threaded  and  has  a 
pitch  diameter  of  2£  ins.  Determine  efficiency  of  this  gear.  Assume  cast- 
iron  worm  wheel  and  steel  worm,  n  =  .05. 


CHAPTER  XII 
ROPE  TRANSMISSION 

TEXTILE  ROPES 

TEXTILE  ropes  are  used  very  extensively  for  the  transmission  of 
power  in  cotton,  steel  and  flouring  mills,  both  for  main  drives  and 
for  individual  machines.  For  the  transmission  of  medium  and 
large  powers,  from  250  H.P.  up,  and  where  the  distance  between 
shafts  is  considerable,  rope  drives  possess  many  advantages. 
Among  these  may  be  mentioned  economy  in  first  cost  and  main- 
tenance, noiseless  and  steady  running,  small  space  required  on 
shaft  by  rope  sheaves  and  ease  with  which  power  may  be  divided 
and  transmitted  to  various  floors  of  a  building. 

Rope  Materials.  Cotton,  hemp  and  manila  hemp  are  the 
materials  most  widely  used  in  the  manufacture  of  transmission 
rope.  Cotton  makes  the  most  flexible  rope  and  is  therefore  better 
adapted  for  driving  individual  machines  where  centers  between 
shafts  are  short  and  lack  of  space  prohibits  the  use  of  large  rope 
pulleys.  Manila  fiber  is  considerably  stronger  than  cotton  and  is 
largely  used  for  heavy  main  drives.  The  rope  material  is  laid 
into  strands  and  these  are  twisted  together  to  form  +he  rope. 
Three,  four,  and  sometimes  six  strands  are  used  for  transmission 
rope.  Figs.  12-1  and  12-2  show  sections  of  such  ropes.  It  will 
be  noted  that  the  four-strand  rope  is  laid  about  a  central  core  and 
is  of  a  more  nearly  circular  section ;  it  will  therefore  give  somewhat 
better  surface  wear  than  the  three-strand  rope.  The  sectional 
area  of  this  rope  is  approximately  .64d2. 

Systems  of  Rope  Driving.  There  are  two  systems  of  rope 
driving,  known  as  the  multiple,  or  English,  and  the  continuous,  or 
American  system.  Although  engineers  differ  widely  as  to  the 
relative  merits  of  these  systems,  there  can  be  no  doubt  that  each 
has  its  field  of  usefulness.  In  the  multiple  system  a  number  of 
independent  ropes  are  used  side  by  side,  sufficient  to  transmit 

121 


122 


ELEMENTS  OP  MACHINE  DESIGN 


the  required  power.  In  the  continuous  system  a  single  endless 
rope  passes  from  the  first  groove  of  the  driver  to  the  first  groove 
of  the  driven  back  to  the  second  groove  of  driver,  and  so  on 
continuously  to  the  last  groove  of  driven  from  which  it  is  returned 


FIG.  12-1. 


FIG.  12-2. 


to  first  groove  of  driver  by  means  of  one  or  more  guide  pulleys 
One  of  these  guide  pulleys  is  placed  in  a  weighted  movable  frame 
called  a  tension  carriage,  by  means  of  which  sufficient  tension  is 
put  on  the  ropes  to  prevent  their  slipping  in  grooves  of  pulleys 
when  transmitting  power. 


FIG.  12-3. 

The  multiple  system  is  especially  adapted  for  large  main 
drives  if  horizontal  or  nearly  so.  The  difficulty  is  to  get  each  rope 
to  transmit  its  proportionate  part  of  the  entire  load.  To  do  this 


ROPE  TRANSMISSION 


123 


the  grooves  must  be  of  the  same  diameter,  the  ropes  must  be  uni- 
formly spliced  and  be  put  on  with  the  same  tension.  With  sudden 
variations  in  load  the  ropes  tends  to  jump  out  of  their  grooves. 
To  prevent  this  many  engineers  use  deeper  grooves  than  for  the 

QFensfon 


FIG.  12-4. 


continuous  system.  There  is  greater  freedom  from  breakdowns 
with  this  system,  as  the  breaking  of  one  rope  does  not  cause 
shut  down.  It  is  also  somewhat  cheaper  to  install.  The  con- 
tinuous system  is  much  used  for  small  and  medium  powers, 
for  driving  individual  machines,  and  for  exposed  drives.  This 


124 


ELEMENTS  OF  MACHINE  DESIGN 


system  is  distinguished  by  its  great  flexibility,  as  it  easily  adapts 
itself  to  difficult  conditions  such  as  vertical  drives  and  quarter 
twist  drives.  The  tension  carriage,  if  properly  installed,  main- 
tains at  all  times  a  uniform  tension  on  each  turn  of  the  rope. 
It  automatically  takes  up  the  stretch  in  the  rope  and  therefore 
changes  in  atmospheric  conditions  do  not  affect  such  a  drive. 
Fig.  12-3  shows  the  distribution  of  power  in  a  mill  by  means  of 
the  multiple  system,  while  Fig.  12-4  shows  a  vertical  drive  using 
the  continuous  system. 

Rope  Pulleys.  In  order  to  secure  sufficient  adhesion  for  driv- 
ing the  rope  pulley  is  provided  with  grooves.  Many  different 
forms  of  grooves  have  been  used  in  which  the  angle  between 
the  sides  varies  from  30°  to  75°.  Fig.  12-5  shows  a  groove 


r 


FIG.  12—5. 


FIG.  12-6. 


which  gives  average  American  practice.  The  proportions  are 
p  =  lid+i",  s  =  fd,  h  =  l%d  and  a  =  45°.  This  value  of  the  angle 
a  is  well-nigh  universal  for  ordinary  conditions  of  driving.  As 
no  wedging  action  is  desired  in  guide  pulleys  the  grooves  of  these 
are  made  so  that  the  rope  rests  on  its  bottom  as  shown  in  Fig. 
12-6. 

Since  the  greatest  wear  on  a  rope  is  internal,  due  to  the  rubbing 
of  the  fibers  over  each  other  as  the  rope  is  bent  around  a  pulley, 
the  diameter  of  these  should  be  as  large  as  practicable.  It  is 
customary  to  make  the  minimum  pulley  diameter  for  cotton 
ropes  thirty  times  the  rope  diameter,  and  for  manila  forty  times 
the  rope  diameter.  The  external  wear  of  rope  is  due  to  slipping. 
To  minimize  this  the  grooves  must  be  turned  accurately  to  the 
same  diameter  and  outline.  They  should  be  finished  smooth  by 


ROPE  TRANSMISSION 


125 


polishing.     As  rope  pulleys  run  at  a  comparatively  high  rate  of 
speed  they  should  be  carefully  balanced. 

Power  Transmission.  The  driving 
capacity  of  a  rope  like  that  of  a 
belt  depends  upon  the  friction  between 
rope  and  pulley  and  therefore  the 
method  of  deducing  the  power  trans- 
mitted is  similar  to  that  of  Chapter 
IX.  Since  the  rope  lies  in  a  groove  a 
radial  force  Q  (Fig.  12-7)  produces  a 
normal  pressure,  Nt  between  the  rope 
and  the  sides  of  groove,  this  pressure 
is  given  by 


Q 


sm  a 


and  the  friction  is 


sin- a 


(2) 


that  is,  the  effect  of  the  groove  is  to  increase  the  value  of  the 


coefficient  of  friction  from      to 


sin  a 


If  the  symbol,  ju',  be  sub- 


stituted for  this  we  have  just  as  in  equation  (3)  of  Chapter  IX, 


r 

22 


(3) 


The  value  of  //  may  conservatively  be  taken  at  .4,  which  gives 


£  =  3.5. 


(4) 


Since  ropes  run  at  comparatively  high  speeds  the  effect  of 
centrifugal  force  cannot  be  neglected. 
Let    W  =  weight  of  1  in.  of  rope, 

V  =  velocity  in  feet  per  second, 
g  =  32. 2  =  acceleration  of  gravity, 
C  =  centrifugal  force  per  inch  of  rope, 
Ct  =  tension  in  rope  due  to  centrifugal  force, 
R  =  radius  of  pulley  in  inches; 


126  ELEMENTS  OF  MACHINE  DESIGN 

then 

w*. 

R' 
912 
and 

C,-™* (5) 

y 

The  weight  of  transmission  rope  may  be  taken  at  .34d2  per 
foot,  this  gives 

C«  =  .01Q5dV (6) 

Using  the  same  notation  as  for  belts,  we  have 

Ti-T2  =  P; 

but  the  maximum  tension  which  occurs  is 

T=Ti+Ct; 

and  therefore 

Ti  =  T-Ct, 

T-(T2+Ct)=P (7) 

The  average  breaking  strength  of  manila  rope  is  TOOOd2  and 

of  cotton  rope  5000d2.     In  order  to  give  long  life  to  transmission 

rope  its  factor  of  safety  is  very  high,  thus  the  value  of  the  maximum 

allowable  stress,  T,  for  rrlanila  rope  is  200d2  and  for  cotton  180d2. 

Let     H  =  horse-power  transmitted  by  one  rope, 

Vm  =  velocity  of  rope  in  feet  per  minute; 
then 

PVm 


ff  = 


33000 

(Tl-T2)Vm 
33000      ' 


(8) 


Substituting  for  T2  its  value  ^\  from  equation  (4), 

»  o.O 

5  TiVm 
H  =  7  33000' 
but  since 


m 
~7  {      33000          ' 


ROPE  TRANSMISSION 


127 


For  manila  rope  if  the  values  given  above  for  T  and  Ct  be  suk 
stituted  this  equation  becomes 


33000 


46200 


(10) 


EXAMPLE.  How  many  2-in.  ropes  are  required  to  transmit 
500  H.P.  at  a  rope  speed  of  3600  ft.  per  minute. 

From  equation  (10)  the  horse-power  transmitted  by  one 
rope  is 


H 


(200  -  .0105  X  3600)4  X  3600 


46200 


:50  nearly; 


2 


\ 


and  therefore  the  number  of  ropes  necessary  is  -^f  =  10. 

If  equation  (10)  be  plotted  it  will  be  noticed  that  the  maximum 
horse-power  is  reached  at  a  speed  of  about  5000  ft.  per  minute 
and  beyond  this  speed  the  power 
decreases,  due  to  the  effect  of 
centrifugal  force.  In  Fig.  12-8 
the  equation  has  been  plotted 
for  a  2-in.  rope. 

Wire-rope  Transmission.  The 
use  of  wire  rope  for  the  trans- 
mission of  power  alone  has  a 
limited  field  to-day,  electric 
transmission  having  taken  its 
place;  but  for  hoisting  and 
conveying  of  material  it  finds 
very  extensive  application.  The 
materials  used  for  the  wires 
of  these  ropes  are  Swedish  iron 
or  steel.  The  steel  is  a  high- 
grade  cast  steel,  the  tensile  strength  of  which  in  the  form 
of  wire  is  from  150,000  to  250,000  Ibs.  per  square  inch.  The 
rope  commonly  used  for  power  transmission  consists  of  six 
strands  laid  about  a  hemp  or  wire  core.  Each  strand  contains 
Seven  or  nineteen  wires,  as  shown  in  Figs.  12-9  and  12-10.  Th^ 


0    1000  2000  3000  1000  5000  6000  7000  8000 
Feet  per  Hour 


FIG.  12-8. 


128 


ELEMENTS  OF   MACHINE  DESIGN 


diameter  of  the  wires  used  depends  of  course  on  the  number  of 
wires  in  each  strand  and  may  be  taken  approximately  at  one- 


FIG.  12-9. 


ninth  rope  diameter  for  the  seven-wire  strands  and  one-fifteenth 
rope  diameter   for  the  nine  teen-wire  strand.     As  the  flexibility 


FIG.  12-10. 

of  a  rope  depends  on  the  size  of  the  wires  used,  it  is  evident  that 
the  nineteen-wire  strands  give  a  more  pliable  rope  and  where  great 

flexibility  is  desired,  a  thirty-seven- 
wire  strand  is  sometimes  used. 

Wire-rope  Pulleys.  These  are 
usually  made  of  cast  iron  and  as 
light  as  consistent  with  considera- 
tions of  strength.  Large  pulleys 
may  have  wrought-iron  arms  set 
into  cast-iron  hubs  and  rim.  They 
are  grooved  as  in  textile-rope  trans- 
missions; but  unlike  such  ropes  the 
wire  rope  is  not  wedged  in  between 
the  sides  of  the  groove,  as  that 
would  cause  rapid  destruction. 
The  rope  rests  on  bottom  of 
groove  and  a  filling  of  some  soft 
material  is  used,  such  as  rubber, 
wood  or  leather  (Fig.  12-11).  The 
FIG  12-11  diameter  of  these  pulleys  is  com- 

paratively large  in  order  to  reduce 

the  stress  due  to  bending  around  the  pulley.     The  ratio  of  pulley 
diameter,  D,  to  rope  diameter,  d,  may  be  taken  as  follows:   for 


ROPE  TRANSMISSION  129 

steel  rope  7  wires  to  the  strand  -^  =  80,   for   19  wire  strand   —, 

cL  d 

=  50;  for  iron  wire  these  values  should  be  doubled. 

Transmission  of  Power.  The  minimum  distance  between 
shafts  is  about  60  ft.  The  maximum  distance  depends  on  the 
topography  of  the  ground,  for  horizontal  transmission  it  is  about 
600  ft.  Across  a  river  or  valley  this  distance  may  be  very  much 
greater.  For  very  long  distances  intermediate  carrying  sheaves 
are  used,  or  a  series  of  spans  are  employed,  the  intermediate 
pulleys  having  two  grooves.  If  the  transmission  is  not  in  a  straight 
line  guide  pulleys  or  bevel  gears  may  be  used  at  the  angles. 

In  wire  transmission  rope  the  stresses  are,  first,  that  due  to 
the  weight  of  rope  (sw),  second,  that  due  to  bending  of  rope  around 
pulleys  (s»)  and  third,  that  the  due  to  centrifugal  force  (sc). 
The  sum  of  these  stresses  must  not  exceed  the  safe  stress  (s)  of 
the  rope.  It  is  only  the  first  of  these  which  produces  adhesioti 
of  the  rope  to  the  pulley  and  therefore  is  a  measure  of  the  tran& 
mission  capacity  of  the  rope.  We  then  have 


and,  using  the  same  notation  as  for  textile  ropes 


Ti-T2  =  P. 

For  a  rope  making  a  half  lap  around  a  rubber  or  leather-filled 
pulley,  slightly  greasy,  the  coefficient  of  friction,  /*,  may  be 
taken  at  .23,  substituting  this  value  we  obtain 


The  adhesive  tension,  Ti,  is  due  to  the  weight  of  the  rope  and 
depends  on  the  curve  in  which  it  hangs. 


where  5  =  diameter  of  the  individual  wires  and  n  is  the  number 
of  wires  in  the  rope. 


130  ELEMENTS  OF   MACHINE  DESIGN 

The  stress  due  to  bending  of  rope  is,  according  to  Bach 


where  E  is  the  modulus  of  elasticity.  The  centrifugal  stress  may 
be  obtained  similarly  to  that  of  belts.  The  weight  of  ordinary 
wire  rope  with  hemp  core  is  about  1.6d2  Ibs.  per  foot;  therefore 


Va  is  the  speed  of  rope  in  feet  per  second.  This  stress  is  quite 
small  in  comparison  to  sw  and  s&,  it  may  in  most  cases  therefore 
be  neglected  without  serious  error.  The  safe  stress,  s,  should  not 
exceed  one-seventh  of  the  breaking  strength.  The  following 
example  will  show  the  applications  of  these  equations. 

EXAMPLE.  What  horse-power  may  be  transmitted  by  a 
J-in.  rope  (steel)  six  19-wire  strands.  The  sheaves  are  5  ft. 
diameter  and  the  rope  runs  at  4200  ft.  per  minute. 

S»=i4=!x29000000x3lr 

=  6050  Ibs.  per  square  inch, 
sc  =  .  13  VS2  =  .  13X4900  =  640  Ibs.  per  square  inch. 

The  safe  stress  s  may  be  taken  at  25000  Ibs.  per  square  inch, 
then 

Su>  =  s—  (sb+sc)  =  17,860  Ibs.  per  square  inch. 

The  rope  contains  114  wires  of  about  -^  in.  diameter,  then 
Ti=    (A)2  XI  14X17860  =1780  Ibs. 


and 


PV  _  890X4200 
"  33000  ~     33000 


Deflection  of  Ropes.    The  towers  which  support  the  sheaves  in 
a  long-distance  transmission  must  be  sufficiently  high  so  that  rope 


ROPE  TRANSMISSION 


131 


may  clear  the  ground.  The  curve  in  which  rope  hangs  is  approxi- 
mately a  catenary,  but  it  will  greatly  simplify  calculation  and  intro- 
duce no  appreciable  error  if  a  parabola  be  substituted.  Then  the 
sag  (Fig.  12-12)  is 

L2W 


where 


h= 

&1 

h  =  sag  in  feet, 
L  =  span  in  feet, 

W  =  weight  in  pounds  of  1  ft.  of  rope, 
T= tension  in  rope. 


FIG.  12-12. 

In  the  above  example  the  sag  in  driving  side  of  rope,  if  a  span 
of  300  ft.  be  assumed,  is 


for  slack  side  of  rope 


3002X1.6 
4X8X1780 


^.O    IL.J 


300JX1.6 


=  5  ft. 


4X8X890 

If  the  points  of  support  are  not  in  the  same  horizontal  plane 
then  the  following  equations  will  give  the  amount  and  position 
of  maximum  sag  as  in  Fig.  12-13. 


__ 
Ll~ 


LTD 


2    LW 
L     TD 


. 

">!= 


WLi* 
2T  ' 

WL22 
2T  * 


132 


ELEMENTS  OF    MACHINE  DESIGN 


The  tension  T  is  that  at  the  lowest  point  of  rope.  At  any  other 
point  x  the  tension  is  T  plus  the  weight  of  a  piece  of  rope  of 
length  hy  or 

Tx=T+hxw. 


FIG.  12-13. 


The  following  table  gives  the  horse-power  which  steel  rope  may 
safely  transmit  as  recommended  by  the  Trenton  Iron  Co.  This 
is  for  wood-filled  pulleys  having  approximately  the  diameters 
previously  given. 


TABLE  OF  HORSE-POWER 


- 

VELOCITY  OF  ROPE  IN  FEET  PER  SECOND. 

10 

20 

30 

40 

50 

60 

70 

80 

90 

100 

1 

4 

8 

13 

17 

21 

25 

28 

32 

37 

40* 

A 

7 

13 

20 

26 

33 

40 

44 

51 

57 

62 

1 

10 

19 

28 

38 

47 

56 

64 

73 

80 

89 

A 

13 

26 

38 

51 

63 

75 

88 

99 

109 

121 

i 

17 

34 

51 

67 

83 

99 

115 

130 

144 

159 

A 

22 

43 

65 

86 

106 

128 

147 

167 

184 

203 

I 

27 

53 

79 

104 

130 

155 

179 

203 

225 

247 

tt 

32 

63 

95 

126 

157 

186 

217 

245 

1 

38 

76 

103 

150 

186 

223 

1 

52 

104 

156 

206 

i 

68 

135 

202 

ROPE  TRANSMISSION  133 

PROBLEMS 

1.  An  engine  developing  2000  H.P.  at  110  R.P.M.  has  a  rope  wheel  12  ft. 
in  diameter.     Determine  how  many  2-in.  manila  ropes  are  required  for  this 
transmission. 

2.  In  a  certain  factory  3200  H.P.  was  to  be  distributed  to  four  floors  as 
follows:    1st  floor  1600  H.P.,  2d  floor  700  H.P.,  3d  floor  500  H.P.,  and  4th 
floor  400  H.P.;    IHn.  ropes  were  used.     The  rope  wheel  of  engine  is  18-ft. 
in  diameter  and  runs  at  90  R.P.M.     Find  the  number  required  for  each 
floor. 

3.  An  engine  developing  600  H.P.  at  140  R.P.M.  has  a  rope  wheel  10  ft. 
in  diameter.     It  is  connected  to  a  jack  shaft,  running  at  220  R.P.M.,  by  means 
of  2-in.  ropes.     From  the  jack  shaft  the  power  is  transmitted  by  l£-m.  ropes 
as  follows:   280  H.P.  to  first  floor,  140  H.P.  to  second  floor,  and  180  H.P.  to 
third  floor;   these  ropes  having  a  velocity  of  3800  ft.  per  min.     Find  numb^T 
of  ropes  required  for  each  drive. 

4.  A  f-in.  diameter  steel  wire  rope  connects  two  sheaves  60  ins.  in  diameter 
and  200  ft.  between  centers,  the  velocity  of  rope  being  3600  ft.  per  minute. 
The  rope  has  six  7-wire  strands,  the  wire  diameter  being  .08  in.     Find  the 
horse-power  which  can  be  transmitted. 


CHAPTER  XIII 
CHAIN  GEARING 

THE  transmission  of  power  by  means  of  a  chain  finds  its 
most  extensive  application  in  motor  cars  and  in  the  individual 
driving  of  machine  tools  by  means  of  electric  motors.  It  is 
largely  due  to  the  bicycle  that  transmission  chains  have  reached 


-§£  35ESE  35)- 


J     L 


FIG.  13-1. 

their  present  state  of  perfection.  The  types  of  chain  used  for 
power  transmission  are  the  block  chain  (Fig.  13-1),  the  roller 
chain  (Fig.  13-2),  and  the  so-called  high-speed  or  inverted  tooth 
chains,  of  which  Fig.  13-3  shows  the  Renold  silent  chain.  These 

f 


\\^JJ             IS 

m           vw; 

rf 

j,                    ,  i.    il 

-fol  n  —  n 

FIG.  13-2. 

chains  may  be  used  on  very  short  as  well  as  on  fairly  large  center 
distances.  They  give  positive  speed  ratio  and  minimum  pressure 
on  bearing,  as  the  slack  side  is  under  no  tension.  A  speed  ratio 
of  one  to  seven  or  one  to  eight  may  easily  be  obtained  with  $ 
single  pair  of  sprocket  wheels. 

134 


CHAIN  GEARING  135 

The  maximum  speed  of  block  chains  is  about  800  ft.  per  min- 
ute, of  roller  chains  1000  ft.  per  minute,  and  the  inverted  tooth 
chain  1500  ft.  per  minute.  While  these  speeds  are  exceeded  in 
some  cases  they  represent  a  safe  limit  with  long  life  of  chain  and 
sprockets.  Chains  of  short  pitch  run  more  smoothly  and  are 
preferable  to  those  of  long  pitch.  It  is  desirable  to  have  the 
distances  between  centers  of  shafts,  connected  by  chains,  ad- 


"i — r 


FIG.  13-3. 

justable.     This  permits  the  inevitable  stretch  of  chain  to  be  taken 
up. 

Sprocket  wheels  must  be  accurately  cut  in  order  to  insure 
quiet  running  and  long  life  to  chain.  The  important  dimensions 
are  outside  diameter,  pitch  diameter,  and  root  diameter.  For 
block  chains 

Let  N  =  number  of  teeth, 

6  =  diameter  of  round  part  of  block, 
C  =  center  to  center  in  block, 
A  =  center  to  center  in  side  links, 

180 

N 

then  (Fig.  13-4), 

sin  a 
tan 


<*=  N> 


C  . 
•j+cos  a 


pitch  diameter  =  D  =  - — - , 
sin  |8 

root  diameter  =  Di  =  D  —  b. 


136  ELEMENTS  OF  MACHINE  DESIGN 

For  roller  chain  (Fig.  13-5)  let 

d  =  diameter  of  roller, 
180 

a~  N' 

p  =  pitch  of  chain; 
then 


sin  a 
Dt  =  D-d, 


for  very  small  sprockets  the  outside  diameter  is  reduced  slightly. 

M^  _/^ 


FIG.  13-4. 


FIG.  13-5. 


The  chief  source  of  trouble  in  chain  drives  is  the  elongation 
of  chain.  This  causes  it  to  bind  on  sprocket  wheel  and  results 
in  rapid  destruction.  In  the  inverted  tooth  type  this  trouble 
is  obviated,  the  effect  of  elongation  being  simply  to  make  the 
chain  ride  up  higher  on  teeth  of  gears.  The  parts  of  trans- 
mission chain  are  made  of  high-grade  steel,  not  so  much  for  its 
greater  strength,  but  for  its  wearing  qualities  when  hardened  and 
tempered.  The  tension  which  may  be  put  on  a  chain  is  limited 
by  the  allowable  pressure  on  pins  or  rivets.  Experience  has 
shown  that  this  pressure  should  not  exceed  600  or  700  Ibs.  per 
square  inch  or  projected  area. 


CHAPTER  XIV 
PIPES  AND  CYLINDERS 

Material  and  Manufacture.  The  metals  used  in  the  manu- 
facture of  pipes  are  cast  iron,  wrought  iron,  steel,  copper,  brass, 
lead,  tin,  etc.,  the  first  three  being  by  far  the  most  important  in 
engineering  work.  Cast-iron  pipe  is  used  chiefly  for  water  and 
other  liquids  as  well  as  steam  when  the  pressure  does  not  exceed 
100  Ibs.  per  square  inch.  Wrought-iron  and  steel  pipes  are 
either  butt  or  lap  welded.  The  small  size  pipes  up  to  about 


FIG.  14-1.  FIG.  14-2. 

2£  ins.  diameter  are  generally  butt  welded  while  above  that  size 
the  lap  welding  process  is  employed. 

Butt-welded  pipe  is  made  by  drawing  the  plate,  called  skelp, 
through  a  bell-shaped  die  (Fig.  14-1)  which  bends  it  up  and 
makes  the  weld.  The  plate  having  of  course  been  previously 
heated  in  a  gas  furnace  to  a  welding  temperature.  In  the  lap- 
welded  tube  the  edges  of  the  plate  are  first  scarfed  or  bevelled  and 
it  is  then  bent  into  the  shape  shown  in  Fig.  14-2.  It  is  then 
reheated  to  a  welding  temperature  and  passed  over  an  arbor 
between  a  pair  of  rolls  (Fig.  14-3)  which  make  the  weld. 

A  newer  process  makes  seamless  tubing.  The  method  of 
manufacture  is  to  pierce  a  billet  of  steel  and  force  it  over  an 
arbor  between  two  rolls.  The  axes  of  these  rolls  are  not  quite 
parallel,  which  gives  the  billet  a  rotary  as  well  as  a  forward 
motion,  thus  drawing  it  out  into  a  tube. 

137 


138 


ELEMENTS  OF   MACHINE  DESIGN 


Pipes  and  cylinders  of  very  large  diameter  are  built  up  by 
riveting  together  steel  or  iron  plate. 


FIG.  14-3. 

Strength  of  Pipes  and  Cylinders.  A  pipe  of  length  L  and  diam- 
eter D  (Fig.  14-4)  is  subjected  to  an  internal  fluid  pressure  of 
p  Ibs.  per  square  inch.  If  we  imagine  any  plane  as  AB  passed 
through  the  axis  of  the  pipe  the  component  (R)  of  the  pressure, 


FIG.  14-5. 


acting  perpendicular  to  this  plane  produces  tension  in  the  sections 
cut  by  it  from  the  pipe.  Since  the  total  tension  produced  must 
be  equal  to  the  force  producing  it,  then  if  t  is  the  thickness  of  pipe 
and  st  the  stress 

R  =  2tLst U) 

Since  R  is  the  sum  of  the  vertical  components  of  the  internal 
pressure  its  value  is  given  by  the  equation. 

R  =  DLP; (2)* 

*  See  Appendix  C  for  derivation. 


PIPES  AND  CYLINDERS 

then  from  (1)  and  (2), 

DLP  =  2tLst; 
and  therefore 

,    Dp 


139 


(3) 


In  practice  this  equation  is  modified  to  allow  for  various  con- 
tingencies such  as  shock  due  to  water  hammer  and  to  handling 
while  in  transportation,  stresses  due  to  method  of  supporting 
pipe,  etc.  This  is  allowed  for  by  adding  a  certain  amount  to 
the  thickness  of  pipe  and  equation  (3)  becomes 


(4) 


k  is  a  constant  the  value  of  which  depends  on  the  material  of 
the  pipe  and  on  the  methods  of  manufacture.  Table  14-1  gives 
average  values  of  this  constant  and  of  st. 

In  built-up  cylinders  and  pipes  the  weakest  portion  is  the 
joint  and  its  efficiency  will  determine  the  thickness  of  plate 
required.  For  this  case  then  if  e  is  the  efficiency  of  the  joint 
expressed  decimally  equation  (3)  becomes 


Dp 


(5) 


TABLE  14-1 


k 

In. 

St 

Cast-iron  pipe  

} 

4000 

Welded-steel  pipe  „  

A 

20,000 

Seamless  steel  pipe  

0 

40,000 

Copper  pipe           .        

t 

8,000 

Cast-iron  engine  cylinders        .    .        

1 

3  000 

Cast-iron  pump  cylinder 

i 

3  000 

The  stress  in  a  transverse  section  of  a  pipe  or  cylinder  is 
one-half  that  in  the  longitudinal  section.  In  Fig.  14-6  is  shown  a 
cylinder  closed  at  the  ends.  Pass  a  plane  AB  through  it  per- 
pendicular to  the  axis  it  will  cut  a  ring  section  the  area  of  which 


140 


ELEMENTS  OF   MACHINE  DESIGN 


is  very  nearly  irDt  (if  D  is  large  in  comparison  with  t).  The 
total  stress  in  this  section  is  equal  to  the  pressure  on  end  of 
cylinder,  or 

7T 


and  therefore 


Dp 


(6) 


1 


FIG.  14-6. 


FIG.  14-7. 


FIG.  14-8. 

Pipe  Joints.  There  are  many  different  joints  used  for  con- 
necting the  lengths  of  pipes.  For  cast-iron  gas  or  water  pipe, 
laid  under  ground,  the  bell-and-spigot  joint  shown  in  Fig.  14-7 
is  commonly  used.  To  make  a  tight  joint  it  is  first  packed  with 
jute  or  hemp  and  then  lead  is  poured  into  the  remaining  space. 
Cast-iron  steam  pipe  is  flanged  as  shown  in  Fig.  14-8,  A  gasket 


PIPES  AND  CYLINDERS 


141 


of  some  soft  material  such  as  rubber,  asbestos,  etc.,  is  used  and 
the  flanges  are  bolted  together.  The  table  below  gives  dimensions 
of  cast-iron  pipe  and  flanges  as  adopted  by  the  Amer.  Soc.  of 
Mech.  Engs.  and  the  Master  Steam  and  Hot  Water  Fitters 
Assoc. 

TABLE  2 


Pipe  Size 
in 
Inches. 

Thickness. 

Diameter 
of 
Flange. 

Thickness 
of 
Flange. 

Diameter 
of 
Bolt  Circle. 

Number 
of 
Bolts. 

Diameter 
of 
Bolts. 

2 

.409 

6 

I 

41 

4 

! 

2* 

.429 

7 

H 

5^ 

4 

f 

3 

.448 

7^ 

1 

6 

4 

i 

3^ 

.466 

81 

H 

7 

4 

I 

4 

.486 

9 

H 

7^ 

4 

i 

4^ 

.498 

91 

tt 

7| 

8 

1 

5 

.525 

10 

H 

81 

8 

1 

6 

.563 

11 

i 

9* 

8 

1 

7 

.600 

12* 

1A 

10| 

8 

1 

8 

.639 

13| 

H 

HI 

8 

1 

9 

.678 

15 

11 

131 

12 

1 

10 

.713 

16 

i& 

141 

12 

i 

12 

.790 

19 

a 

17 

12 

I 

14 

.864 

21 

U 

18| 

12 

i 

15 

.904 

22i 

U 

20 

16 

i 

16 

.946 

23£ 

l& 

211 

16 

i 

18 

.02 

25 

l& 

221 

16 

U 

20 

.09 

27^ 

itt 

25 

20 

U 

22 

.18 

29| 

at 

27i 

20 

11 

24 

.25 

32 

u 

29£ 

20 

U 

26 

.30 

34i 

2 

31f 

24 

11 

28 

1.38 

36£ 

2^ 

34 

28 

11 

30 

1.48 

38| 

2* 

36 

28 

11 

36 

1.71 

45| 

2| 

421 

32 

U 

42 

1.87 

52| 

2| 

49£ 

36 

H 

48 

2.17 

59| 

2! 

56 

44 

li 

For  pipe  carrying  up  to  200  Ibs.  per  square  inch  pressure. 


142 


ELEMENTS  OF   MACHINE  DESIGN 


Wrought-iron  and  steel  pipes  find  a  wide  application  in  various 
fields  and  the  joints  used  are  very  numerous.     For  small  pipe  and 


FIG.  14-9. 


FIG.  14-10. 


FIG.  14-11. 


FIG.  14-12. 


FIG.  14-13. 


low  pressures  a  coupling  as  in  Fig.  14-9  is  used.  For  high  pres. 
sures  and  large  pipe  some  form  of  flanged  joint  is  used.  Fig 
14-10  shows  a  screwed  flange  joint.  Fig.  14-11  a  joint  in  whict 


PIPES  AND  CYLINDERS  143 

the  flanges  are  fastened  to  pipe  by  expanding  it  into  grooves 
turned  into  the  hub  of  flanges.  To  make  a  tight  joint  the  flanges 
are  provided  with  tongue  and  groove.  Fig.  14-12  shows  a  joint 
with  loose  bolt  flanges,  and  rings  welded  on,  which  are  provided 
with  a  turned  groove  for  a  copper  wire  ring  packing.  These  last 
three  are  examples  of  the  type  of  joint  used  for  high  pressures  and 
superheated  steam.  Fig.  14-13  is  a  joint  used  for  small  pipes 
called  a  union.  This  provides  an  easy  method  of  taking  pipes 
apart  where  it  may  be  necessary  occasionally. 


Fia.  14-14. 

Expansion  Joints.  Long  lines  of  pipe  handling  hot  fluids 
expand  to  such  an  extent  that  provision  must  be  made  for  this 
if  leaky  joints  or  breaking  of  pipe  is  to  be  prevented.  This  is 
particularly  true  with  the  high-pressure  and  highly  superheated 
steam  used  in  present  day  steam  engineering  practice.  The 
expansion  per  100  ft.  of  pipe  carrying  steam  at  100  Ibs.  pressure 
will  be  about  2J  ins.  With  high  pressure  and  superheat  it  will 
be  about  double  this  amount.  The  simplest  way  of  providing 
for  the  necessary  flexibility  is  by  means  of  expansion  bends,  one 
form  of  which  is  shown  in  Fig.  14-14.  These  may  be  used  for 


144 


ELEMENTS  OF  MACHINE  DESIGN 


pipes  up  to  12  ins.  diameter.  The  radius  R  should  not  be  less 
than  five  times  the  diameter  of  pipe.  Fig.  14-15  shows  a  type  of 
expansion  joint  with  gland  and  stuffing  box,  perhaps  the  most 
satisfactory  way  of  taking  care  of  the  expansion  in  long  lines  of 
pipe. 

Support  for  Pipe  Lines.  Expansion  causes  motion  of  the  entire 
line  of  pipes.  Therefore,  in  order  to  have  the  expansion  joints 
perform  their  duty,  it  is  essential  to  provide  fixed  points  at  which 
the  pipe  is  firmly  anchored ;  between  two  such  pionts  an  expansion 
joint  is  placed.  Besides  these  the  pipe  must  of  course  have  fre- 


FIG.  14-15. 

quent  points  of  support  to  prevent  undue  bending  stresses.  Fig. 
14-16  shows  a  roller  support  and  bracket  for  large  pipes.  Fig. 
14-17  is  an  example  of  one  of  the  many  types  of  overhead  supports 
or  hangers. 

Thick  Cylinders.  For  very  high  pressures  such  as  are  common 
in  hydraulic  presses,  the  cylinders  are  very  thick  and  it  is  found 
that  the  stress  is  no  longer  uniformly  distributed.  It  is  a  maxi- 
mum on  the  inside  and  decreases  toward  the  outside.  Since 
equation  (3)  is  based  on  the  assumption  of  uniform  stress  it  is 
not  applicable  to  this  case.  Bach  has  investigated  this  subject 
experimentally  and  has  deduced  an  approximate  equation  as 
follows : 

Let  D  =  outside  diameter  of  cylinder, 
d  =  inside  diameter  of  cylinder; 


PIPES  AND   CYLINDERS 


145 


then 


but 

and  therefore 


FIG.  14-16. 

Size  of  Pipe.  It  is  frequently  necessary  to  determine  the  size 
of  a  pipe  to  deliver  a  given  quantity  of  a  fluid  when  the  velocity 
of  the  fluid  is  known. 

Let    d  =  inside  diameter  of  pipe  in  inches, 

A  =  internal  sectional  area  in  square  inches, 
V  =  velocity  of  fluid  in  feet  per  min., 
Q  =  quantity  of  fluid  to  be  delivered  in  cubic   feet   per 
minute; 


146 
then 

and 


ELEMENTS  OF  MACHINE  DESIGN 


O=i=.« 

V     144 ' 


and 


/.    Q  = 


576 


d  = 


(8) 


FIG.  14^17. 

Example.  What  should  be  the  internal  diameter  of  a  pipe 
which  delivers  the  feed  water  to  a  boiler  requiring  a  maximum 
of  5000  Ibs.  per  hour,  the  velocity  of  water  in  pipe  to  be  150  ft. 
per  min. 

Since  water  weighs  approximately  62.4  Ibs.  per  cubic  foot, 

5000 


PIPES  AND  CYLINDERS  147 

and 

d  =  13. 5* 


PROBLEMS 

1.  The  thickness  of  a  standard  wrought  iron  pipe  10  ins.  diameter  is  .366 
in.     This  pipe  is  tested  at  factory  to  a  pressure  of  250  Ibs.  per  square  inch. 
Determine  stress  in  pipe  during  test. 

2.  Find  thickness  of  steam-engine  cylinder  if  its  diameter  is  15  ins.  and  it 
operates  with  steam  at  125  Ibs.  per  square  inch. 

3.  Find  thickness  of  shell  for  a  boiler  60  ins.  in  diameter.     The  steam 
pressure  is  100  Ibs.  per  square  inch.     The  longitudinal  seam  is  a  double-riveted 
butt  joint  having  an  efficiency  of  75  per  cent. 

4.  A  hydraulic  press  is  to  exert  a  pressure  of  200,000  Ibs.  with  a  pressure  of 
1000  Ibs.  per  square  inch  in  the  cylinder.     Determine  thickness  of  cylinder 
assuming  it  to  be  of  cast  iron  and  allowing  a  tensile  stress  of  3000  Ibs.  per  square 
inch. 

5.  A  heating  boiler  evaporates  1500  Ibs.  of  water  per  hour,  the  volume  of 
the  steam  being  20  cu.ft.  per  pound.     Determine  the  size  of  the  steam  main 
so  that  the  steam  velocity  will  be  2000  ft.  per  minute. 


CHAPTER  XV 
VALVES 

Valves  are  machine  parts  to  regulate  the  flow  or  the  pres- 
sure of  fluids  in  pipes  and  other  containers.  They  may  be  divided 
into  two  general  classes:  (a)  those  which  have  no  regular  periodic 
motion  and  usually  are  operated  by  hand,  or,  in  some  cases,  by 
the  pressure  of  the  contained  fluid;  (6)  those  having  a  regular 
periodic  motion  generally  obtained  from  the  machine  of  which  they 
are  a  part,  as  a  steam  engine  valve,  or  due  to  the  periodically 
varying  pressure,  as  in  a  pump. 

The  surface  with  which  the  valve  is  in  contact  when  closed 
is  called  the  seat  of  the  valve.  Another  classification  is  accord- 
ing to  their  motion,  relative  to  the  valve  seat.  Thus  we  have 
(1)  swing  valves,  often  called  flap  or  clack  valves,  which  rotate 
about  an  axis  parallel  to  the  plane  of  the  valve  seat;  (2)  slide 
valves,  which  slide  parallel  to  the  plane  of  the  valve  seat;  and 
(3)  lift  valves,  which  move  perpendicular  to  the  plane  of  the 
/alve  seat. 

Types  of  Valves.  In  the  following  pages  are  shown  a  few 
examples  of  the  more  important  types  of  valves  belonging  to 
classification  (a)  above.  Fig.  15-1  represents  a  globe  valve. 
This  is  used  on  pipes  up  to  about  4  or  5  ins.  in  diameter.  The  valve 
disc  is  renewable,  the  material  being  copper,  bronze  or  some  softer 
substance,  such  as  vulcanized  fiber  or  rubber.  Where  the  valve 
spindle  passes  through  the  top  of  casing  is  a  receptacle,  called 
stuffing  box,  which  is  filled  with  a  packing  to  prevent  leakage.  . 

If  the  flow  is  in  the  direction  indicated  by  arrow,  then,  when 
valve  is  closed,  the  pressure  being  on  top  of  valve  disc  will  tend 
to  tighten  it  on  its  seat.  A  disadvantage,  however,  is  that  the 
stem  cannot  be  repacked  while  under  pressure. 

In  the  gate  valve  shown  in  Fig.  15-2  the  flow  is  unobstructed 
and  this  type  is  commonly  used  for  the  larger  sizes  of  pipes. 
In  the  very  large  sizes  these  valves  are  frequently  operated  by 

148 


VALVES 


149 


an  electric  motor.  The  hand  wheel  may  have  only  a  turning 
motion  acting  as  a  stationary  nut,  or  it  may  be  attached  to  the 
valve  spindle  and  rise  with  it,  as  in  the  figure. 

The  check  valve  is  used  to  prevent  the  fluid  from  flowing  back 
along  a  pipe  when  the  pressure  is  relieved.     Fig.  15-3  illustrates 


FIG.  16-1. 

one  example  of  this  type.     For  small  pipes  this  valve  frequently 
takes  the  form  of  a  ball  resting  on  a  spherical  valve  seat. 

Safety  Valves.  Vessels  containing  fluids  under  pressure  are 
provided  with  safety  valves.  The  purpose  of  these  is,  as  their 
name  implies,  to  prevent  the  pressure  rising  above  a  safe  limit; 
that  js,  they  open  automatically  whenever  the  pressure  reaches 
a  predetermined  limit.  Their  most  important  application  is  in 
the  steam  boiler.  In  Fig.  15-4  is  shown  such  a  valve,  in  which 


150 


ELEMENTS  OF  MACHINE  DESIGN 


the  pressure  necessary  to  keep  it  closed  is  obtained  by  means  of 
a  lever  and  weight.  When  a  spring  is  used  for  this  purpose  it  is 
called  a  pop  safety  valve. 

Cocks.     In  Fig.  15-5  is  shown  a  simple  cock,  consisting  of  a 
Conical  plug  which  fits  into  a  similar  shaped  opening  in  body  of 


FIG.  15-2. 


valve.    The  angle  a  enclosed  between  sides  of  cone  varies  from 
8°  to  15°. 

The  materials  used  for  these  various  valve  bodies  are  brass, 
cast  iron  and  steel.  Brass  is  used  in  the  small  sizes  for  both 
low  and  high  pressures.  Above  200°  C.  the  strength  of  brass  rap- 
idly decreases,  it  is  therefore  not  suitable  for  superheated  steam. 


VALVES 


151 


Cast  iron  is  used  for  low  pressures  except  in  the  smallest  sizes, 
where  it  is  also  employed  for  high  pressures.  For  very  large  valves 
and  very  high  pressures  steel  castings  are  the  most  practical. 


FIG.  15-4. 

Lift  Valves.  In  Fig.  15-6  are  shown  some  of  the  types  of  lift 
valves  commonly  employed.  Let  h  be  the  lift  of  the  valve, 
then  in  order  to  make  the  valve  opening  equal  to  the  area  under 
valve  we  have  (Fig.  15-6a). 


152 


ELEMENTS  OF   MACHINE  DESIGN 


and  therefore 


(1) 


This  type  of  valve  is  operated  by  the  pressure  of  the  fluid  under 
it  as  in  pumps  and  air  compressors.  In  order  to  secure  quick 
closing  of  valves  they  are  often  loaded  with  a  spring. 


Let 


FIG.  15-6a. 


FIG.  15-66. 


F  [/  =  area  subjected  to  pressure  under  valve, 
FA  =  area  subjected  to  pressure  above  valve, 
pu  =  pressure  per  square  inch  acting  on  Fu, 
PA  =  pressure  per  square  inch  acting  on  FAl 
W  =  weight  of  valve, 

g  =  acceleration  of  gravity, 

a  =  acceleration  of  valve, 

P  =  pressure  of  spring; 


then  for  equilibrium 


—  a. 
y 


.     (2) 


That  is,  the  valve  will  rise  until  the  forces  acting  upwards  and 
downwards  are  balanced  as  shown  by  equation  (2).     As  soon  as 


VALVES 


153 


the  velocity  ot  che  stream  through  valve  becomes  zero,  it  will 
begin  to  close.  But  since  this  does  not  take  place  instantane- 
ously, some  of  the  liquid  will  flow  back,  furthermore,  the  valve 
will  seat  with  an  accelerated  motion  which  causes  hammering  of 


FIG.  15-7. 

valve  and  seat;    for  these  reasons  it  is  desirable  to  make  the  lift 
as  small  as  possible. 

In  very  large  valves  the  openings  are  a  series  of  concentric 
rings,  as  in  Fig.  15-7.  This  permits  large  valve  opening  with 
very  small  lift  of  valve.  Thus  if  d\  and  d2  are  the  outside  and 


154  ELEMENTS  OF  MACHINE  DESIGN 

inside  diameters  of  one  ring,   then,   assuming  equal  velocities 
through  seat  and  valve. 


and 

,     di  —  d,2    b 


That  is,  the  lift  of  valve  is  equal  to  one-half  the  breadth  of  one 
ring  opening;   it  is,  however,  generally  increased  to  .66  or  .76. 

There  are  many  forms  of  valves  for  special  purposes,  which 
can  be  taken  up  profitably  only  in  connection  with  the  machines 
of  which  they  form  a  part,  as  they  cannot  be  treated  independently 
of  the  mechanism  which  operates  them. 


CHAPTER  XVI 
FLY-WHEELS 

THE  purpose  of  a  fly-wheel  is  to  maintain  the  speed  of  a 
machine  which  is  doing  work  or  receiving  energy  at  a  variable 
rate,  between  certain  predetermined  limits.  It  is  a  reservoir 
of  energy,  storing  it  up  when  the  energy  is  supplied  at  a  greater 
rate  than  the  demand  and  giving  it  out  when  the  reverse  is  the 
case.  It  must  be  understood  that  in  order  to  perform  its  func- 
tion there  must  be  some  variation  in  speed;  that  is,  a  fly-wheel 
can  only  store  energy  when  its  speed  is  increasing  and  give  out 
energy  when  its  speed  is  decreasing. 

In  such  machines  as  punches,  shears,  horizontal  presses, 
etc.,  fly-wheels  are  used,  as  work  is  done  during  a  small  portion 
of  the  cycle  only.  If  then  such  a  machine,  without  fly-wheel, 
were  belt  driven,  the  belt  would  have  to  be  large  enough  to  sup- 
ply the  entire  energy  consumed  by  the  work  to  be  done  during 
the  working  portion  of  the  cycle  only,  and  would  run  idle,  except 
for  friction  of  machine,  during  the  remaining  portion  of  cycle. 
By  the  use  of  a  fly-wheel  a  much  smaller  belt  may  be  employed 
which  supplies  energy  at  approximately  a  constant  rate  through- 
out the  cycle. 

Design  of  Fly-wheels.  To  determine  the  weight  of  a  fly- 
wheel for  a  machine,  it  is  first  necessary  to  know  the  amount  of 
energy  to  be  stored  and  the  speed  variation  which  may  be  allowed. 
The  kinetic  energy  in  a  rotating  fly-wheel,  neglecting  the  small 
amount  in  hub  and  arms,  is 

WV2 
K  =  ~HAbs. 

Where  W  is  weight  of  rim  in  pounds,  V  is  the  mean  velocity  of 
rim  in  feet  per  second,  and  g  is  the  acceleration  of  gravity.     If 

155 


156 


ELEMENTS  OF   MACHINE  DESIGN 


the  velocity  of  the  rim  be  changed  from  V  to  V\  the  kinetic  energy 
becomes 


and  therefore  the  energy,  E,  absorbed  or  given  out  during  this 

change  of  speed  is 

W 
E=K-K1  =  ^(V2-V12'); 

from  which  we  obtain 


As  an  illustration  of  the  method  of  finding  weight  of  a  fly-wheel 
take  the  horizontal  press  shown  in  Fig.  16-1.     Here  A  is  the  driv- 


FIG.  16-1. 

ing  shaft  and  B  the  driven  shaft.  Let  A  make  six  revolutions 
to  one  of  B.  The  work  is  done  while  B  makes  one-third  of  a  revolu- 
tion, the  other  two-thirds  being  idle.  Let  the  belt  speed  be  1000 
ft.  per  minute  and  assume  the  actual  work  to  be  done  during  each 
cycle  of  operations  is  8000  ft.-lbs.  Shaft  B  makes  40  R.P.M. 
The  efficiency  of  machine  is  to  be  taken  at  80  per  cent. 
The  horse-power  required  to  drive  this  machine  is 


H  = 


8000X40 
.80 X 33000 : 


12.1. 


FLY-WHEELS  157 

The  effective  belt  pull  is 

12.1X33000 

P=T^-T*=   nrnoo" 

=  4001bs. 

The   distance   the   belt   travels   during   working   portion   of 

1000 
cycle  is  -|X    ,n   =  8^  ft.  and  therefore  the  work  done  by  belt 

during  this  period  is  8fX  400  =  3333  ft.-lbs.     The  total  work  to 

8000 
be  done  during  each  cycle  is  -^-  or  10,000  ft.-lbs.     The  energy 

to  be  stored  during  idle  part  of  cycle  to  be  given  out  in  working 
portion  of  cycle  is 

E=  (10,000  -3333).  80  =  5334  ft.-lbs. 

Let  the  mean  diameter  of  fly-wheel  rim  be  48  in.  and  the  speed 
variation  allowable  10  per  cent.  The  fly-wheel  is  placed  on  shaft 
A  and  therefore  makes  240  R.P.M.  The  mean  rim  speed  when 
running  at  its  greatest  velocity  is 

T7    TTX  48X240 

V=       12X60     -50  ft.  per  second. 

The  mean  rim  speed  when  fly-wheel  runs  at  its  lowest  velocity 
just  after  completing  working  portion  of  cycle  is 

vi  =  .  90X50  =  45  ft.  per  second; 
therefore  from  equation  (1), 


50-45 
=  7301bs. 

Cast  iron  weighs  .26  Ib.  per  cubic  inch,  therefore  the  rim 

730 
must  contain  -^  =  2800  cu.in.     The  length  of  rim  is  7rX48=150 

2800 
ins.;  then  the  sectional  area  of  rim  is  -r-  =  18.7  sq.in.,  or  the  rim 


may  be  3J  by  5f  ins.  as  shown  in  Fig.  16-2. 


158 


ELEMENTS  OF   MACHINE  DESIGN 


Very  often  a  graphical  method  of  design  is  the  simplest. 
This  is  particularly  the  case  when  work  is  done  at  a  variable 
rate  and  sufficient  information  is  obtainable  to  draw  a  work 
diagram.  In  outline  the  process  is  as  follows:  In  Fig.  16-3 
let  the  area  of  A  BCD  represent  the  energy  supplied  to  machine 
per  cycle.  As  this  area  is  a  rectangle  the  energy  is  supplied  at  a 
constant  rate.  The  area  AMORSD  represents  the  work  done. 


FIG.  16-2. 


FIG.  16-3. 


But  the  total  energy  supplied  must  equal  the  total  work  done 
or 

ABC 'D  =  AMORSD. 

Subtracting  the  common  area  AMNPRSD  from  each  side  of 
the  equation  we  have 

MBN+PRSC=NOP. 

From  B  to  N  evidently  the  supply  of  energy  is  greater  than  the 
demand  by  the  amount  represented  by  the  area  MBN  and  this 
surplus  is  stored  in  the  fly-wheel.  From  N  to  P  the  demand 
exceeds  the  supply  by  an  amount  represented  by  area  NOP  and 
the  fly-wheel  gives  out  energy;  finally  from  P  to  C  an  amount 
equal  to  PRSC  is  again  stored  in  fly-wheel.  In  other  words, 
the  diagram  gives  us  the  energy  E=  NOP,  which  the  fly-wheel 
must  be  capable  of  storing  up.  If  now  its  mean  speed  and  the 
amount  of  variation  permissible  be  known,  its  weight  may  be 
calculated  from  equation  (1). 


FLY-WHEELS 


159 


Coefficient  of  Fluctuation.  The  difference  of  maximum  and 
minimum  speed,  in  revolutions  per  minute,  divided  by  the  mean 
speed  is  called  the  coefficient  of  fluctuation,  so  that  if 

JV  =  mean  fly-wheel  speed  in  R.P.M., 
Ni=  maximum  fly-wheel  speed  in  R.P.M., 
N2  =  minimum  fly-wheel  speed  in  R.P.M. 

the  coefficient  of  fluctuation  is 

6  ==• 


N 
Below  are  given  usual  values  of  6  for  various  kinds  of  machinery 


Shearing  and  punching  machines 06  - .  04 

Pumps 05-. 03 

Flour-mill  machinery 04  - .  03 

Paper-mill  machinery  and  machine  tools  .    .  03  - .  025 

Spinning  machinery 02 -.01 

Dynamos 01  -.003 

Stresses   in   Fly-wheels.     The  stresses  which  occur    in    the 
rim  and  arms  of  a  fly-wheel  are  very  complicated  and  no  solution 


FIG.  16-4. 


entirely  satisfactory  has  been  arrived  at.     In  general  the  rotation 
of  the  wheel  produces  centrifugal  forces  in  both  the  rim  and  the 


160 


ELEMENTS  OF   MACHINE  DESIGN 


arms,  these  in  turn  produce  tensile  stresses  in  both.  But  since 
the  arms  and  rim  are  rigidly  fastened  together  and  do  not  stretch 
equal  amounts  the  effect  is  to  cause  bending  stresses  in  rim,  as 
indicated  in  Fig.  16-4.  Speed  variations  still  further  complicate 
the  problem  as  the  inertia  forces  caused  thereby  act  tangentially 
and  produce  bending  stresses  in  the  arms. 

The  tensile  stress  in  a  rotating  iron  ring  due  to  centrifugal 
force  may  be  found  exactly  in  the  same  way  as  was  done  for 
leather  belting  in  Chapter  IX,  and  the  same  equation  holds  good, 
that  is 

=  12WV2 

Where 

ct  —  tensile  stress  per  sq.in.  due  to  centrifugal  force, 
W  —  weight  of  one  cu.in.  of  cast  iron, 
V  =  velocity  of  rim  in  feet  per  second. 

Substituting  numerical  values  for  W  and  g,  we  obtain 


FIG.  16-5. 

If  we  substitute  the  ultimate  tensile  stress  Ut  =  20,000  for 
ct  and  solve  (2)  for  V,  we  obtain  the  speed  at  which  a  cast-iron 
wheel  would  burst  if  subjected  to  centrifugal  stresses  only.  The 


FLY-WHEELS 


161 


result  is  F  =  454  ft.  per  second.  It  must  be  noted  that  this 
ultimate  speed  does  not  depend  on  the  size  or  shape  of  the  rim 
section.  It  is  customary  to  assume  that  a  rim  speed  of  100  ft. 
per  minute  is  the  safe  limit  for  cast-iron  wheels.  For  higher 
speeds  other  materials  are  to  be  preferred. 


FIG.  16-6. 

Construction  of  Fly-wheels.  Small  fly-wheels  up  to  7  or  8  ft. 
diameter  are  commonly  cast  in  one  piece.  With  heavy  rims  and 
hubs  dangerous  casting  strains  may  result.  To  avoid  these  the 
hub  is  split  into  two  or  more  parts,  this  enables  the  arms  to 
contract  freely  when  cooling.  Figs.  16-5  and  16-6  show  two 


FIG.  16-7. 

constructions,  in  the  first  the  hub  is  split  into  two  parts  and  bolted 
together,  while  the  latter  has  hub  split  into  three  parts  and  shrink 
links  fitted  at  each  end.  Large  wheels  are  cast  in  two  or  more 
sections.  Fig.  16-7  shows  a  wheel  cast  in  two  sections  which  are 
held  together  by  bolts  and  shrink  links.  This  joint  is  much  weaker 
than  the  rim  section  besides  causing  heavy  additional  stresses 
due  to  the  centrifugal  force  of  lugs  and  bolts.  If  joint  is  placed 
in  center  line  of  arm  as  in  Fig.  16-8  it  is  generally  stronger  than 


162 


ELEMENTS  OF   MACHINE  DESIGN 


FIG.  16-8. 


.:TfH- 


FIG.  16-9. 


FLY-WHEELS 


163 


the  first  form.  Fig.  16-9  illustrates  a  large  fly-wheel  in  which 
each  arm  together  with  a  portion  of  the  rim  is  cast  separately. 
The  arms  are  bolted  to  a  hub  which  is  cast  in  two  parts.  The  rim 
sections  are  held  together  by  steel  links.  The  rim  is  provided 
with  barring  holes  for  rotating  wheel  by  means  of  a  lever  when 
setting  valves,  etc.  The  next  illustration,  Fig.  16-10  is  of  a  fly- 
wheel in  which  the  rim  and  arms  are  cast  in  one  and  are  then 
bolted  to  the  hub.  The  arms  are  dished,  which  gives  them  greater 
flexibility,  and  probably  makes  a  stronger  fly-wheel. 


FIG.  16-10. 

PROBLEMS 

1.  In  a  certain  machine  the  mean  fly-wheel  diameter  is  72  ins.     Weight  of 
wheel  1400  Ibs.     The  maximum  and  minimum  rim  speeds  are  175  and  160 
R.P.M.  respectively.     Determine  the  change  in  kinetic  energy  of  fly-wheel. 

2.  Determine  weight  of  fly-wheel  required  for  a  punch-press  for  following 
conditions.     Diameter  of  wheel  (mean)  54  ins.     Maximum  speed  240  R.P.M. , 
minimum  speed  216  R.P.M.     The  machine  requires  18  H.P.     The  complete 
cycle  consists  of  eight  revolutions  of  fly-wheel,  only  three  of  which  take 
place  during  working  portion  of  cycle.     There  are  28  complete  cycles  per 
minute. 

3.  A  fly-wheel  14  ft.  in  diameter  rotates  at  125  R.P.M.     The  fly-wheel 
rim,  width  16  ins.  and  depth  =20  ins.,  is  cast  in  two  parts  held  together  by  two 
shrink  links  at  each  joint*     Determine  dimensions  of  section  of  these  links 
if  a  stress  of  20,000  Ibs.  per  square  inch  is  allowed. 


CHAPTER  XVII 


CRANK-SHAFTS,  CRANK-PINS,  AND  ECCENTRICS 

Stresses  in  Crank   Shafts.     In  Fig.  17-1  is  shown  diagram- 
^fl.m  engine  mechanism  and  the  forces  acting  on 


stresses  in  ^raiLK  onaiis.  in  rig.  i/— i  is  siiown  uiagram- 
itically  the  steam  engine  mechanism  and  the  forces  acting  on 
;  various  machine  elements.  The  steam  pressure  P  on  piston  is 

-F. 


ma 

the  various 


FIG.  17-1. 

transmitted  through  the  piston  rod  to  the  cross-head  pin.  Here 
it  is  resolved  into  a  thrust  T,  in  the  connecting  rod  and  a 
pressure  R  on  the  guides.  The  thrust,  T,  is  transmitted  by 
connecting  rod  to  crank-pin,  where  it  may  be  resolved  into  a 
tangential  component  T\  and  a  force  Q,  causing  either  com- 
pression or  tension  in  crank-arm. 


FIG.  17-2. 

Taking  first  the  crank-shaft,  as  shown  in  Fig.  17-2,  having 
the  crank  at  one  end  and  carrying  the  fly-wheel  between  the  two 
bearings,  let 

G  =  weight  of  fly-wheel, 

P  =  total  thrust  on  crank-pin   (assumed   equal  to  steam 
pressure  on  piston). 
164 


CRANK-SHAFTS,   CRANK-PINS,   AND  ECCENTRICS       165 

The  shaft  is  subjected  to  combined  twisting  and  bending.  For 
convenience  it  may  be  assumed,  without  serious  error,  that  this 
pressure  on  the  crank-pin  acts  horizontally  for  all  positions  of 

IG 


FIG.  17-3. 

crank.     To  find  the  reactions  at  A  and  B  find  first  the  reactions 
at  these  supports  in  the  horizontal  plane  (Fig.  17-3). 


RAH= 


„         Ph. 

KBH  =  --  , 


and  then  in  vertical  plane 


L  * 

The  total  reactions  are 

- CD 

: (2) 

The  bending  moments  at  A  and  C  may  now  be  determined.     At  A 

MA  =  Ph (3) 

AtC 

Mc  =  RBh. (4) 

The  maximum  twisting  moment  transmitted  is 
T=Pr, 


166  ELEMENTS  OF   MACHINE  DESIGN 

The  equivalent  twisting  moment  at  A  is 

TA=MA+VT*+MA* (5) 

AtC  

Tc  =  Mc+^/T2-\-M(?;        (6) 

then  the  diameter  at  A  is 

and  at  C 

3  \T 

(8) 

The  value  of  s  for  a  mild  steel  shaft  may  be  taken  at  from 
7000  to  9000  Ibs.  per  square  inch.  It  is  customary  in  designing 
a  crank-shaft  to  assume  tentative  dimensions  and  use  the  above 
equations  (1)  to  (8)  to  check  the  stress  at  sections  A  and  C. 

EXAMPLE.  The  following  data  are  taken  from  a  small  hori- 
zontal engine: 

P=  10,000  Ibs., 
G  =  2500  Ibs., 
r  =  8  ins., 
dA  =  5  ins., 
dc  =  6  ins., 
Zi=9  ins., 
£2  =  28  ins., 
Z3  =  24ins. 
Find  stress  at  A  and  C. 

10000X61 


AH= 


7nn  , 
=  11,700  Ibs., 


10000X9     17nnlK 
RBH  =  -  ^  -  =  1700  Ibs. 
OA 

2500X24 
RAV  =  -    -  =  H50  Ibs., 


2500X28     1QF:niK 
RBV  =  -  ^  —  =  1350  Ibs., 


2  =  11,750  Ibs., 


=  V(1700)2+(1350)2  =  2170  Ibs., 


CRANK-SHAFTS,   CRANK-PINS,   AND  ECCENTRICS       167 


MA  =  9  X  10,000  =  90,000  in.-lbs., 
Mc  =  2170X24  =  52,500  in.-lbs. 
T  =  10,000  X  8  =  80,000  in.-lbs., 


TA  =  90,000 + A/800002 + 900002 
=  210,000  in.-lbs., 


Tc  =  52500+ V  800002+525002 

=  148,000  in.-lbs. 

Then  the  stress  at  section  A  is  (from  equation  (7)) 
5.1X210000 


125 


=  8550  Ibs.  per  square  inch; 


and  at  section  C 


5.1X148000     oeo/,«  •     , 

s= =  3580  Ibs.  per  square  inch. 


Cranked-shafts.     This  type  of  crank-shaft  is  used  in  horizontal 
engines  having  the  forked  frame,  called  center-crank  engine,  and 


; 

zZ 

- 

p 

1 

[G. 

<  — 

17-4. 

in  nearly  all  vertical  engines;  such  a  shaft  is  shown  in  Fig.  17-4. 
In  small  engines  the  shaft  is  forged  out  solid,  while  for  large 


FIG.  17-5. 

multiple-cylinder  engines  it  is  built  up  as  in  Fig.  17-5.     Here 
the  shaft,  crank-cheeks  and  crank-pins  are  forged  separately  and 


168 


ELEMENTS  OF  MACHINE  DESIGN 


fastened  together  by  means  of  shrmk  or  force  fits,  a  key  being 
used  on  each  crank  as  an  additional  safeguard. 

Calculation  of  Cranked  Shaft.  Let  Fig.  17-6  represent  in 
outline  the  crank-shaft  of  a  vertical  engine.  In  this  case  the  steam 
pressure  on  piston  and  the  weight  of  fly-wheel  act  in  the  same 
direction.  Then 

Pk-Gh 


RA  = 


RB  = 


Ph+G(L+h) 


At  B  the  bending  moment  is 
MB  =  ( 
|G 


(ID 


FIG.  17-6. 
and  the  twisting  moment  is 


The  equivalent  twisting  moment  is 


and  the  diameter  of  shaft  at  B  is 


FIG.  17-7. 


(12) 
(13) 

(14) 


The  diameter  at  A  is  usually  made  equal  to  that  at  B,  or  the 
bearing  is  designed  so  as  to  give  equal  bearing  pressure  at  A 
and  B. 

The  bending  stress  at  C  will  be  a  maximum  on  the  upward 
stroke  as  in  Fig.  17-7.     For  this  position 

_Pk+Gh.  (.. 

**4   *  r  ,  \10) 


CRANK-SHAFTS,   CRANK-PINS,   AND  ECCENTRICS       169 

and  the  bending  moment  at  C  is 

Mc  =  RAl (16) 

The  twisting  moment  is  again 

T  =  Pr (17) 

The  equivalent  twisting  moment  for  this  section  is 

(18) 


the  diameter  at  C  is 

3  / — 

cfc-1.72^,       (19) 

or 

s  =  ^ •     (20) 

It  is  better  to  design  this  part  of  shaft  for  bearing  pressure  and 
use  equation  (20)  as  a  check  to  keep  stress  within  safe  limits. 

Taking  the  crank  cheek,  the  maximum  stress  will  be  at  the 
section  h,  where  it  joins  the  shaft.  The  bending  moment  at  this 
section  is 

M,  =  Pr; (21) 

and  the  twisting  moment  is 

Th=  RA(li~{-l4)—Ph)      ....     (22) 
and  the  equivalent  bending  moment  is 


',    .     .     .     (23) 
then  if  b  is  breadth  and  t  is  thickness  of  cheek  (see  Fig.  17-4) 

(24) 


or 


In  the  above  calculations  it  has  been  assumed  that  the  maxi- 
mum stress  will  occur  when  crank  is  at  right  angles  to  line  of 
cross-head  motion. 


170 


ELEMENTS  OF  MACHINE  DESIGN 


Cranks.  In  side-crank  engines  where  an  overhung  crank-pin 
is  used  the  crank  is  made  separate  from  the  crank-shaft  to  which 
it  is  fastened  by  a  force  or  shrink  fit  with  the  additional  safe- 
guard of  a  pin  key.  Fig.  17-8  is  an  example  of  a  forged  crank. 


FIG.  17-8. 

The  material  used  is  generally  steel,  sometimes  wrought  iron; 
cast  iron  only  for  small,  cheaply  constructed  engines.  In  Fig. 
17-9  is  shown  a  cast-steel  crank. 


FIG.  17-9. 

The  stresses  to  which  the  crank-arm  is  subjected  are  com- 
bined bending  and  torsion.  Thus  in  Fig.  17-8  the  bending 
moment  at  any  section  distant  x  from  crank  pin  is 


the  twisting  moment,  which  is  the  same  at  all  sections,  is 
T  =  Pa. 

If  we  imagine  the  sides  extended  to  center  line  of  crank-shaft 
the  maximum  bending  moment  will  occur  at  this  section  and  is 

M  =  Pr-f 


CRANK-SHAFTS,   CRANK-PINS,   AND  ECCENTRICS 
and  the  equivalent  bending  moment  is 


then 
and 


171 


(26) 


The  stress  s  for  wrought  iron  should  not  exceed  7000  Ibs.  per  square 
inch  and  for  steel  9000  Ibs.  per  square  inch.  In  designing  the 
crank-arm  the  distance  a  is  not  definitely  known  and  must  be 
assumed.  The  first  calculations  will  give  simply  tentative  values 
which  are  used  for  recalculation  if  the 
assumed  dimensions  have  been  found  wide  of 
the  mark.  The  length,  7,  of  the  boss  is  from 
.9d  to  1.2d,  its  thickness,  t,  is  from  Ad  to  .5d. 
In  calculating  b  and  h  from  equations 
(26)  we  obtain  the  dimensions  of  a  solid 
rectangle.  The  actual  section  at  center  of 
crank-shaft  is  shown  in  Fig.  17-10,  and  after  FIG.  17-10. 

having  chosen    values    of   t   and    I  as  given 
above,  the  section  modulus  of  this  section   must   be   at   least 
equal  to  that  of  the  solid  rectangle,  or 


EXAMPLE.  In  an  engine  having  a  cylinder  20  ins.  diameter 
and  taking  steam  at  100  Ibs.  per  square  inch  the  crank  has  the 
following  dimensions  (see  Fig.  17-8):  r=15  ins.,  a  =  5j  ins.,  6  =  3J 
ins.,  1  =  8  ins.,  £  =  3J  ins.,  h  =  l2  ins.,  and  d  =  8  ins.  What  is  the 
maximum  stress  in  crank?  Total  pressure  on  crank-pin  is 

P  =  ^X202X100  =  31,400  Ibs. 

Maximum  bending  moment  is 

M  =  31400X15  =  471,000  in.-lbs. 
Twisting  moment  is 

in.-lbs, 


172  ELEMENTS  OF   MACHINE  DESIGN 

The  equivalent  bending  moment  is 

Me  =  }(471000+  V471000*+l7256(?) 
=  493000  in.-lbs. 

The  section  modulus  of  the  dangerous  section  is 


and  therefore  the  stress  is 

_Me_  493000 
>S~  z  "      215 


=  2290  Ibs.  per  square  inch. 


In  high-speed  engines  the  inertia  of  the  reciprocating  parts 
may  increase  the  pressure  on  crank-pin  considerably.  This 
can  be  allowed  for  by  increasing  the  factor  of  safety.  The  same 
thing  should  be  done  if  the  load  on  engines  varies  frequently  and 
suddenly. 

Crank-pins.  Crank-pins  are  essentially  journals,  and  the 
important  consideration  in  their  design  is  to  prevent  heating  by 
keeping  the  bearing  pressure  within  proper  limits.  The  stresses 
to  which  the  pin  is  subjected,  while  of  secondary  importance, 


FIG.  17-11. 


FIG.  17-12. 


should  in  all  cases  be  calculated  to  make  sure  they  are  not  above 
a  safe  value  for  the  material,  which  is  always  steel.  In  some  cases 
of  severe  service  a  special  high-carbon  or  alloy  steel  is  used,  with 
the  bearing  surface  hardened  and  ground.  In  Fig.  17-11  are  shown 
some  of  the  various  methods  of  fastening  the  pin  to  the  crank- 
arm. 

Calculation  of  Crank-pin.     Let  the  allowable  bearing  pressure 
per  square  inch  be  q  ancl  the  dimensions  as  shown  in  Fig.  17-12, 


CRANK-SHAFTS,   CRANK-PINS,  AND  ECCENTRICS       173 

Then  since  the  total  bearing  pressure  must  be  equal  to  the  pres- 
sure on  crank-pin, 

P  =  qld (29) 

Let  the  ratio  of  the  length  of  pin  to  its  diameter  be  denoted  by 
n,  or 

l__ 

d~ 

substituting  this  value  of  /  in  (29)  we  obtain 

(30) 

nq 

The  following  are  usual  values  of  n:  For  high-speed  engines 
n  —  .S  to  1.25;  for  medium-speed  engines  n=?  1  to  1.4  and  for 
low-speed  engines  n=1.25  to  1.6.  The  values  of  q  may  be  taken 
from  subjoined  table. 

Type  of  Machine.  Value  of  q. 

Crank  pins    of  slow-speed  machines  with  intermittent  load. 

(Shears,  punches,  etc.) 2000-3000 

High-speed  stationary  engines 400-700 

Medium-  and  low-speed  engines 600-900 

Gas  engines 300-600 

Locomotives 1000-1800 

Marine  engines 400-600 


The  overhung  crank-pin  is  a  cantilever;  the  maximum  bending 
moment  is 

Pl 


The  moment  of  resistance  of  the  circular  section  is  ^ 
therefore 

Pl     * 


and 


then 


(31) 


174 


ELEMENTS  OF  MACHINE  DESIGN 


Careful  lubrication  is  of  prime  importance  for  continuous 
service.  In  small  engines  an  oil  cup  or  other  lubricating  device 
is  placed  on  the  connecting  rod  end.  In  large  engines  the  lubri- 
cant is  generally  fed  through  a  hole  in  center  of  crank-pin.  In 
Fig.  17-13  is  shown  the  crank-shaft  of  a  large  double-acting  tan- 


(X 

1  1 

A  *.- 

TT 

-'.;- 

1 

rr 

T} 

f  5 

"*- 

1  1 

ii: 

cflai-flr, 

—  y-  —  y- 

-;£;-;£: 

!  i 
xi 

—  t 

FIG.  17-13. 

dem  gas  engine.     The  lubricant  flows  through  the  shaft  supply- 
ing both  main  bearings  and  crank-pins. 

Eccentrics.  The  purpose  of  an  eccentric  is  to  change  rotary 
motion  into  reciprocating,  its  action  being  identical  to  that  of  a 
crank.  In  fact,  it  may  be  regarded  as  a  crank  mechanism  in 
which  the  crank-pin  has  been  increased  in  diameter  until  it  includes 
the  crank-shaft.  Their  use  is  limited  generally  to  such  cases 


FIG.  17-14. 

where  the  crank  radius,  here  called  eccentricity,  is  small  and  the 
resistance  to  be  overcome  not  too  heavy;  their  advantage  being 
that  they  may  readily  be  fastened  at  any  point  of  a  shaft  and  are 
a  much  cheaper  construction  than  a  cranked  shaft. 

The  eccentric  sheave  consists  of  a  disc  either  solid  or  split 
into  two  parts.  The  solid  sheave  can,  of  course,  be  used  only 
where  it  can  be  put  on  over  end  of  shaft.  Fig.  17-14  shows  the 


CRANK-SHAFTS,  CRANK-PINS,  AND  ECCENTRICS        175 


construction  of  such  a  sheave.     The  outside  diameter  of  sheave 


s 


=  2(r+e+t), 


(32) 


where  r  is  radius  of  shaft,  e  the  eccentricity,  and  t  the  minimum 
radial  thickness  of  sheave.  The  thickness,  t,  may  be  obtained 
from  the  empirical  equation 


n. 


The  material  used  for  these  sheaves  is  nearly  always  cast  iron, 
in  marine  engines  steel  is  also  used. 

The  split  sheave  is  shown  in  Fig.  17-15.     Here  the  two  parts 
are  fastened  together  by  studs  and  nuts.     Instead  of  the  nuts, 


FIG.  17-15. 


FIG.  17-16. 


cotters  may  be  used.  In  Fig.  17-16  the  usual  rim  sections  are 
shown.  The  width  b  of  sheave  is  determined  by  the  force  P  on 
eccentric  rod  and  the  rubbing  speed  of  the  contact  surfaces.  Just 
as  in  the  case  of  crank-pins,  we  have  (Fig.  17-12) 

bDq-P; 
and 


Where  q  is  again  the  allowable  pressure  per  square  inch  of  pro- 
jected area,  its  value  depends  on  v,  the  rubbing  velocity  in  feet 
per  minute;  it  may  be  taken  at 

3000C)       50000 
=-     -to-     —  . 


176 


ELEMENTS  OF  MACHINE  DESIGN 


In  steam  engine  work  it  is  often  impossible  to  determine  the  force 
P  with  any  degree  of  accuracy,  and  the  following  value  of  b  will 


FIG.  17-17. 


be  found  to  give  results  agreeing  closely  with  stationary  engine 
practice : 

b  =  .756+1  in. 

Eccentric  straps  are  always  made  in  two  parts,  fastened 
together  by  bolts,  the  material  being 
generally  cast  iron  or,  for  large  engines, 
steel.  In  Fig.  17-17  is  shown  a  common 
form  of  strap.  For  low  speeds  and 
light  pressures  the  contact  surfaces 
may  be  cast  iron,  otherwise  the  strap 
is  lined  with  white  metal.  The 
eccentric  rod  may  be  screwed  into 
strap,  or  it  can  be  fastened  by  means 
of  bolts  as  shown,  or  cotters.  The 
strap  is  designed  as  a  beam  with  a 
FIG.  17-18.  uniformly  distributed  load-  From 


CRANK-SHAFTS,   CRANK-PINS,  AND  ECCENTRICS        177 
Fig.  17-18  the  maximum  bending  moment  at  section  A  B  is 


The  section  modulus  of  the  section  is 


assuming  it  to  be  a  rectangle  of  breadth  b  and  depth  t,  and 
therefore  the  maximum  stress  is 


(34) 


For  cast  iron  s  may  be  taken  at  2500  to  4000  and  for  steel  6000 
to  9000.  The  strap  bolts  are  designed  for  tension,  each  sus- 
taining a  tensile  force  equal  to  \P. 

PROBLEMS 

V 

1.  A  high-speed  engine  has  a  cylinder  12  ins.  in  diameter  and  uses  steam 

at  125  Ibs.  per  square  inch.  The  crank-pin  is  of  the  overhung  type.  Make 
a  suitable  design  of  this  pin,  allowing  a  bearing  pressure  of  550  Ibs.  per  square 
inch.  Check  design  for  stress. 

2.  A  14  by  12-in.  engine  operating  with  steam  at  100  Ibs.  per  square  inch 
has  a  crank-pin,  overhung  type,  5  ins.  diameter  by  4  ins.  long.     Determine 
bearing  pressure  and  stress  in  pin. 

/""  3.  Sketch  shows  the  crank-arm  of  a  14  by  24-in.  engine  operating  with 
steam  at  125  Ibs.  per  square  inch.  Determine  the  stress  in  section  AB  of 
crank-arm. 


CHAP.  XVII.     Prob.  3, 


178 


ELEMENTS  OF  MACHINE  DESIGN 


4.  Sketch  shows  the  crank-shaft  of  a  tandem  compound  steam  engine. 
The  maximum  pressure  on  crank-pin  is  50,000  Ibs.  The  weight  of  fly-wheel 
is  18,000  Ibs.  Determine  stress  in  shaft  at  sections  AB  and  CD. 


—    T 

v 

! 

c 

so'-  

I 

^ 

4  i 

CHAP.  XVII.     Prob.  4. 


5.  The  sheave  of  a  steam  engine  eccentric  is  10  ins.  diameter.  The  load 
is  2000  Ibs.  Determine  breadth  of  eccentric  if  engine  runs  at  200  R.P.M. 
Also  design  strap  similar  to  Fig.  17-17. 


CHAPTER  XVIII 

CONNECTING   RODS,   PISTON   RODS,  AND 
ECCENTRIC    RODS 

THE  connecting  rod  is  used  to  change  the  rectilinear  motion  of 
the  cross-head  into  the  rotary  motion  of  the  crank.  The  material 
used  for  these  rods  is  wrought  iron  or  steel,  occasionally  malleable 
cast  iron.  The  rod  consists  of  two  essential  parts,  the  ends  which 
fit  around  the  crank-  and  wrist-pins,  and  the  shaft  or  rod  con- 
necting these  ends. 

The  rod  is  either  circular,  rectangular  or  I-shaped  in  section, 
as  shown  in  Fig.  18-1.  The  round  section  is  commonly  used 


FIG.  18-1. 

in  long-stroke  slow-speed  engines,  while  locomotives  and  the 
short-stroke  high-speed  engines  more  frequently  show  the  rectan- 
gular section. 

Connecting  rod  ends  show  numerous  variations  in  construc- 
tion. In  general  there  are  two  types,  viz.:  the  solid,  or  closed 
end,  and  the  open  end.  Fig.  18-2  shows  an  example  of  the  closed 
type  for  the  wrist-pin  end.  The  boxes  are  of  brass.  They  are 
adjusted  for  wear  by  means  of  a  wedge  or  cotter.  Another 
form  of  solid  end  is  shown  in  Fig.  18-3.  It  will  be  noticed  that 
the  crank-pin  boxes,  which  are  cast  steel,  lined  with  babbitt 
metal,  are  adjusted  for  wear  by  a  wedge  which,  in  this  case,  moves 
at  right  angles  to  that  of  the  previous  figure.  The  closed  end 
can,  of  course,  be  used  only  where  it  is  possible  to  assemble  it 
over  the  end  of  pin  around  which  it  fits;  thus  it  cannot  be  used 
on  center  crank  engines  or  where  the  crank  pin  is  enlarged  at  the 
end  to  form  a  flange. 

179 


180 


ELEMENTS  OF  MACHINE  DESIGN 


FIG.  18-2. 


FIG.  18-3. 


CONNECTING,   PISTON,   AND   ECCENTRIC  RODS         181 

Fig.  18-4  is  an  open  end  called  the  marine  type.  This  is 
a  favorite  design  for  all  kinds  and  sizes  of  engines  and  other 
machinery.  The  cap  is  held  by  two  bolts  which  must  be  of  suf- 


FIG.  18-4. 

ficient  strength  to  resist  the  entire  pull  due  to  the  steam  pressure 
upon  the  piston.  The  nuts  are  locked  by  means  of  set  screws. 
Another  open  end,  known  as  the  strap  end,  is  shown  in  Fig.  18-5. 
Here  the  boxes  are  held  by  a  steel  or  wrought  iron  strap  which  is 


FIG.  18-5. 

fastened  to  the  stub  end  of  connecting  rod  by  bolts,  as  shown, 
or  a  cotter  may  replace  the  bolts,  as  in  Fig.  18-6. 

Connecting  Rod  Boxes.     In  the  smaller  sizes  the  boxes  are 
either  brass  or  bronze.    For  large  pins  they  are  brass  or  steel  cast- 


182 


ELEMENTS  OF   MACHINE   DESIGN 


ing,  lined  with  babbitt.     The  thickness  at  center  of  box  (Fig* 
18-7)  may  be 

in. 


At  sides  this  is  generally  reduced  to  ti  =  %t  to  f  t,  the  length  between 
the  flanges  is  about 

s  =  .8L. 


=tr 

FIG.  18-6 


FIG.  18-7. 


In  taking  up  the  wear  of  the  boxes  it  should  be  noticed  that  if 
the  adjusting  wedge  is  on  the  inside  brass  at  both  ends  the  dis- 
tance between  wrist-  and  crank-pins  will  be  increased,  while,  if 
placed  on  the  outside  brasses,  rod  is  shortened.  To  maintain 
this  distance  constant  some  engine  builders  place  one  wedge  on 
inside  and  one  on  outside  brass. 

Design  of  Rod.  The  rod  is  subjected  to  alternate  tension 
and  compression.  While  under  the  compression  load  it  acts  as  a 
strut  or  column,  and  as  this  will  produce  the  maximum  stress  in 
rod  it  must  be  designed  as  such.  The  length  of  connecting  rod 
for  stationary  engines  is  commonly  five  times  the  crank  length. 
Where  the  engine  must  be  kept  short,  as  in  vertical  marine  engines, 
this  length  is  about  four  times  the  crank  length,  in  locomotives; 
where  there  is  plenty  of  room  the  connecting  rod  is  from  six  to 
eight  times  the  crank  length. 

It  will  be  convenient  to  assume  a  circular  section  for  the  rod 
first  and  obtain  the  diameter.  An  equivalent  rectangular  section 
may  then  be  obtained,  as  will  be  shown  later.  In  general  the 
connecting  rod  may  be  considered  a  column  with  pin  ends.  If 
the  length  of  rod  is  greater  than  thirty  times  its  diameter  the 
JCuler  equation  given  in  Table  4  of  Chapter  I  will  apply.  As  the 


CONNECTING,  PISTON,  AND  ECCENTRIC  RODS        183 

rod  is  subjected  to  alternate  tension  and  compression  a  high 
factor  of  safety  should  be  used,  say  7^  =  9  to  15. 

The  maximum  thrust  P  will  come  on  connecting  rod  when  the 
crank  is  perpendicular  to  line  of  motion  of  cross-head,  as  in  Fig. 


FIG.  18-8. 

18-8.     If  Q  be  total  pressure  on  piston,  then  from  similar  tri 

angles, 


VL2-R2 
Let  the  ratio  of  connecting  rod  length  to  crank  length  f  •=)  be  n, 


then 


n 


(1) 


Then  using  the  Euler  column  equation  of  Chapter  I 

WEI 


P  = 


FL2' 


Substituting  for  /  its  value  ^r  ^4  we  obtain 

Ed4 


2FL2' 


(2) 


In  using  this  equation  it  is  best  to  assume  a  tentative  value  of 
d  and  then  solve  for  P.     If  the  first  trial  solution  gives  a  value 


184  ELEMENTS  OF  MACHINE  DESIGN 

of  P  too  low  or  considerably  higher  than  the  actual  thrust  on 
rod,  other  assumptions  for  d  are  made  until  a  satisfactory  result 
is  obtained.  With  a  little  practice  it  requires  no  more  than  two 
trials. 

As  generally  the  ratio  of  length  of  connecting  rod  to  diameter 
is  less  than  30  it  will  be  found  that  the  Gordon-Rankine  equa- 
tion is  more  serviceable.  This  for  a  circular  section  is 

*>=—  -72;      .......     (3) 


J 

here  A  is  the  sectional  area  of  the  rod  and  A;  is  a  constant  whose 
value  is  ^^  for  steel  or  wrought  iron.  If  we  substitute  this 
value  equation  (3)  becomes 


EXAMPLE.  Find  the  diameter  of  a  round  connecting  rod,  the 
total  thrust  on  which  is  36,000  Ibs.  ;  the  rod  is  60  ins.  long. 

Make  a  tentative  solution  by  assuming  a  diameter  and  solve 
for  the  safe  thrust  P.  Assume  d  =  2  ins.  and  for  a  steel  rod 
s  =  8000  Ibs.,  then  from  equation  (4), 

8000X^X4 

P=-- 
+ 


750X4 
this  value  is  much  too  low,  therefore  assume  d  =  3,  then 


8000X^X9 


750X9 


This  is  satisfactory  as  the  error  is  on  the  side  of  safety. 

Rectangular  Rods.     In  case  a  rectangular  rod  is  desired  the 
diameter  of  a  round  rod  is  first  calculated,  then  if  t  is  the  thick- 


CONNECTING,   PISTON,   AND  ECCENTRIC  RODS        185 


ness  of  rod  (Fig.  18-9),  and  6  is  the  breadth  at  middle,  let  the  ratio 

of  -  be  denoted  by  n.     The  usual  value  of  n  ranges  from  1.5  to  3. 
t 

The  thickness  can  be  found  from  the  diameter  of  round  rod  by 
means  of  the  equation 

(5) 


The  following  table  gives  values  of  t  for  various  values  of  n. 
n=1.5  1.75  •  2.00  2.25  2.50  2.75  3.00 
t=  .79d  .76d  .74d  .72d  ,7ld 


This  type  of  connecting  rod,  when  used  on  high-speed  stationary 

engines,  is  made  to  taper  from  wrist-pin  end 

to  crank-pin  end,  being  largest  at  the  latter. 

This  is  in  order  to  take  care  of  the  bending 

stresses  acting  in  the  plane  of  motion  at  right 

angles  to  the  axis  of  rod.     The  breadth,  6, 

at  cross-head  end  may  be  about  .86  and  at 

crank-end  1.26. 

Connecting-rod  Ends.  The  sides  are  in 
tension  and  the  sectional  area  should  be  such 
as  to  keep  the  stress  st  within  safe  limits.  pIG 

The  factor  of  safety  should  not  be  less  than  8, 
as  the  load  varies  from  zero  to  a  maximum  in  each  revolution 


i     I 


I     i 
j i 


FIG.  18-10. 

of  engine.     Since  there  are  two  areas  bXti  (Fig.  18-10)  to  carry 
the  load,  we  have 


and 


(6) 


186 


ELEMENTS  OF  MACHINE  DESIGN 


The  cross-piece  connecting  the  sides  is  regarded  as  a  beam  loaded 
with  a  uniformly  distributed  load.  From  Fig.  18-10  we  have  the 
bending  moment  at  center 

PI 


and 
therefore 


PI 

8 


(7) 


In  the  marine  end  the  cap  is  designed  as  above.  The  bolts 
are  in  tension  and  sustain  one-half  of  the  total  load,  P,  each. 
The  stress  in  bolts  not  to  exceed  6000  Ibs.  per  square  inch  for  mild 
steel. 


FIG.  18-11. 

The  forked  end  (Fig.  18-11)  has  its  dangerous  section  at  CD. 
This  section  is  subjected  to  a  direct  tension 


CONNECTING,   PISTON,   AND  ECCENTRIC    RODS        187 
and  a  bending  stress 


__ 

Sb  ~      ' 


bh2 


and  therefore  the  total  stress  is 


(8) 


Piston  Rods.  These  rods  transmit  the  steam  pressure  to  the 
cross-head.  For  small  engines  the  rod  is 
screwed  into  position  and  riveted  over,  as 
shown  in  Fig.  18-12.  A  better  method 
is  to  use  a  nut,  as  in  Fig.  18-13  or 
Fig.  18-14. 

The  piston  rod  is  designed  as  a 
column  with  ends  guided  and  therefore 
equation  (4)  will  also  apply  here.  The 
stress  sc  may  be  4000  to  6000  Ibs.  per 
square  inch.  With  very  heavy  pistons 
the  rod  is  also  subjected  to  a  bending 
stress.  This  may  be  allowed  for  by 

using  a  low  value  of  sc.     If  we  assume  Sc  =  5000  and  remember 
that  the  load  on  piston  rod  is 


FlG 


, .      length  of  rod  (L) 

and  call  the  ratio  -p—         —^ — -j-^  =  n.  then 
diameter  of  rod  (a) 

from  equation  (4) 

5000  X^d2 


and 


(9) 


188 


ELEMENTS  OF  MACHINE  DESIGN 


for  any  value  of  n  we  may  write  this  equation 


(10) 


The  following  table  gives  values  of  K  for  various  values  of  n: 


n=        8 
X=.0148 


10 
.0150 


12 
.0154 


14 
.0159 


16 
.0164 


20  25 

.0175     .0191 


EXAMPLE.  An  engine  has  a  cylinder  18  ins.  diameter  and  18- 
in.  stroke,  the  steam  pressure  is  125  Ibs.  per  square  inch.  Length 
of  piston  rod  is  30  ins.  Find  diameter  of  rod. 


FIG.  18-13. 


FIG.  18-14. 


Assume  a  trial  value  of  n=14  say,   then  d  =  -=2.14    ins. 

Ti 

Now  check  by  equation  (10),  then 

^  =  .0159X18X^125 

=  3.20. 
This  gives  a  value  of 


Therefore  our  assumed  value  is  too  high;  taking  a  second  trial 
value  of  n  =  10  we  obtain 


CONNECTING,   PISTON,   AND  ECCENTRIC  RODS         189 
and  checking  by  equation  (10) 


This  is  sufficiently  close  to  our  assumed  value  and  it  may  there- 
fore be  taken  as  the  diameter  of  piston  rod. 


FIG.  18-15. 

Eccentric  Rods.  These  are  usually  round  except  in  high- 
speed engines,  where  a  rectangular  section  is  frequently  used. 
The  material  is  wrought  iron  or  steel.  If  the  thrust  on  rod  is 


FIG.  18-16. 

known  they  may  be  designed  in  the  same  manner  as  connecting 
rods.     The  calculated  dimensions  may  be  taken  as  those  at  center 


190  ELEMENTS  OF  MACHINE  DESIGN 

of  rod.     Figs.  18-15  and  18-16  show  two  of  the  commonest  types 
of  eccentric  rod  ends. 


PROBLEMS 

sy 

1.  Determine  the  thrust  or  the  connecting  rod  of  a  15  by  16-in.  steam 
engine.     The  steam  pressure  is  160  Ibs.  per  square  inch  and  the  length  of 
connecting  rod  is  32  ins. 

2.  The  total  thrust  on  a  round  connecting  rod  is  50,000  Ibs.     Length  of 
rod  is  72  ins.     Find  diameter  of  rod  if  the  stress  is  7500  Ibs.  per  square  inch. 
%/  3.  The  engine  in  problem   1   has  a  rectangular  connecting  rod.     The 
ratio  of  b:t  is  2.5.     Find  these  dimensions  if  the  safe  stress  is  9000  Ibs.  per 
square  inch. 

4.  In  a  marine   engine  the  low-pressure  cylinder  is  60  ins.  in  diameter; 
the  maximum  pressure  being  40  Ibs.  per  square  inch.     The  piston  rod  is  of 
nickel-steel.     Allowing  a  stress  of  10,000  Ibs.  per  square  inch  find  its  diameter; 
the  length  of  rod  is  72  ins. 

5.  Design  the  crank-pin  end  of  the  connecting  rod  in  problem  2.     Design 
to  be  similar  to  Fig.  18-5.     State  clearly  all  assumptions  and  make  your 
calculations  complete. 


CHAPTER  XIX 


PISTONS,  CROSS-HEADS  AND  STUFFING  BOXES 

PISTONS  are  used  in  engines,  pumps,  air  compressors  and  other 
machines.  Their  purpose  is  to  take  up,  or  to  produce  the  pressure 
of  fluids  enclosed  in  cylinders.  To  do  this  they  must  have  a 
fluid-tight  fit  and  at  the  same  time  move  with  as  little  friction  as 
possible.  This  is  accomplished  by  means  of  packing  rings,  as 
in  Fig.  19-1.  Pistons  which  carry  valves  that  permit  the  fluid 
to  pass  through  them  from  one  side  of  piston  to  the  other  are 
called  buckets  (Fig.  19-2).  If  the  fluid  pressure  acts  on  one  side 


FIG.  19-1. 


FIG.  19-2. 


of  the  piston  only  and  the  packing  is  placed  in  the  cylinder,  as 
in  Fig.  19-3,  the  piston  is  called  a  plunger. 

The  material  most  commonly  used  for  pistons  is  cast  iron. 
In  pumps  where  the  water  to  be  pumped  has  a  corrosive  action 
brass  or  bronze  are  used.  Very  large  pistons,  especially  of 
marine  engines,  are  steel  castings,  in  order  to  reduce  their  weight 
to  a  minimum. 

Engine  Pistons.  Pistons  up  to  about  18  ins.  diameter  are 
usually  of  the  hollow  cast-iron  type  shown  in  Fig.  19-4.  For 
Ordinary  pressures  the  piston  is  provided  with  two  cast-iron  pack- 

191 


192 


ELEMENTS  OF   MACHINE   DESIGN 


ing  rings.     For  very  high  pressures  three  or  more  rings  may  be 
used.     In  small  sizes  the  internal  radial  ri!  s  are  often  omitted. 


FIG.  19-3 


FIG.  19-4. 


The  purpose  of  the  openings  is  to  remove  the  core  after  casting. 
The  proportions  may  be  as  follows: 


=  VD     to     1. 


Large  pistons  are  usually  of  the  built-up  type.  That  is, 
they  consist  of  several  parts  bolted  together.  Figs.  19-5  and 
19-6  show  two  forms  of  this  type  of  piston.  It  will  be  noticed 
that  in  the  built-up  pistons  it  is  not  necessary  to  expand  the 
packing  rings  so  that  they  can  be  pulled  over  the  piston  castings 
as  is  the  case  in  the  solid  pistons.  Fig.  19-7  shows  a  steel  pis- 
ton of  the  type  commonly  used  in  marine  engines. 

In  the  single-acting  gas  engine  the  piston  rod  and  cross-head 
are  omitted,  and  the  side  thrust  of  connecting  rod  is  transmitted 
to  cylinder  walls  through  the  piston.  For  this  reason  the  piston 
is  made  very  long  as  shown  in  Fig.  19-8.  The  length  L  of 
piston  may  be  from  L=1.25D  to  2D,  the  larger  value  being 


PISTONS,   CROSS-HEADS  AND  STUFFING  BOXES         193. 

used  for  small  engines.     The  number  of  packing  rings  is  generally 
from  4  to  8. 


FIG.  19-5. 


FIG.  19-6. 


FIG.  19-7. 


FIG.  19-8. 


Piston  Rings.  Cast  iron  has  been  found  to  be  the  best  material 
for  packing  rings.  These  rings  are  of  two  general  types,  viz.: 
those  in  which  the  pressure  against  the  cylinder  wall  is  due  to 


194 


ELEMENTS  OF   MACHINE  DESIGN 


elastic  forces  within  and  those  in  which  springs  or  other  means  are 
used  for  this  purpose. 

Several  forms  of  the  first  type  are  shown  in  Fig.  19-9.  These 
rings  are  made  so  that  their  diameter  D  is  from  1.08  to  1.12 
times  the  diameter  of  the  cylinder.  A  piece  is  then  cut  out  and 
when  ring  is  placed  in  cylinder  the  ends 
are  sprung  together.  The  elastic  stresses 
thus  produced  press  the  ring  against  the 
cylinder  wall  and  make  a  steam-tight  fit. 
This  pressure  between  rings  and  cylinder  is 
from  2  to  5  Ibs.  per  square  inch  of  contact 
surface.  The  joint  shown  at  a  in  the  figure 
permits  some  leakage  of  the  fluid  under 
pressure,  but  on  account  of  its  simplicity  is 

often  used  in  small  engines.  The  joints  at  6  and  c  to  a  large 
extent  prevent  this  leakage.  The  following  table  gives  good  pro- 
portions for  this  type  of  piston  ring: 

PISTON  RINGS 


s* 

a 

Y 

^->^- 

1 

b 
to.  IS 

-9. 

- 

^" 

X.t 

<r:  j'j 
c 

Diameter. 

Width. 

Thickness. 

Diameter. 

Width. 

Thickness. 

D 

w 

t 

D 

W 

t 

4 

I 

A 

12 

i 

I 

6 

A 

A 

15 

A 

f 

8 

I 

i 

18 

I 

A 

10 

A 

A 

21 

1 

I 

24 

i 

f 

When  the  rings  are  placed  in  piston  the  joints  are  staggered, 
the  rings  being  prevented  from  rotating  by  a  pin  or  screw. 

Cross-heads.  The  cross-head  carries  the  pin  called  wrist-pin 
or  gudgeon-pin  and  connects  the  piston  rod  and  the  connecting 
rod.  It  has  a  reciprocating  straight-line  motion  sliding  on  sur- 
face formed  on  the  guides.  There  are  a  great  variety  of  con- 
structions, the  materials  being  cast  iron,  steel  or  wrought  iron. 

Fig.  19-10  shows  a  simple  form  of  cross-head  for  a  single 
guide,  the  sliding  surfaces  being  planes.  The  piston  rod  is  cot- 
tered  into  the  hub  of  cross-head.  In  Fig.  19-11  is  shown  a  cross- 
head  with  two  cylindrical  sliding  surfaces  formed  on  separate 


PISTONS,   CROSS-HEADS  AND  STUFFING  BOXES         195 

parts  called  slippers.  The  slippers  may  be  of  cast  iron,  brass 
or  cast  iron  with  babbitted  wearing  surfaces.  Another  form  is 
shown  in  Fig.  19-12.  Here  the  slippers  are  arranged  so  that  the 


FIG.  19-10. 


FIG.  19-11. 


FIG.  19-12. 

wear  on  sliding  surfaces  may  be  taken  up.  This  type  is  very 
common  in  American  stationary  engine  practice.  Still  another 
form  of  cross-head  is  shown  in  Fig.  19-13.  Here  the  wrist-pin 


196 


ELEMENTS  OF   MACHINE  DESIGN 


projects  on  each  side  of  cross-head  and  therefore  a  forked  connect- 
ing rod  is  to  be  used. 

Design.  The  cross-head  is  designed  for  bearing  pressure.  The 
maximum  pressure  will  come  on  cross-head  when  crank  has  rotated 
through  90°  from  its  dead  center  position,  as  in  Fig.  19-14.  If 
L  is  length  of  connecting  rod  and  R  length  of  crank,  then,  by  find- 
ing the  components  of  the  steam  pressure  (Q)  on  piston  along  the 


FIG.  19-13. 

<?onnecting  rod  (P)  and  perpendicular  to  the  guides  (G)  we  have 
from  similar  triangles 

G  =  cb=        R  ^ 

Q    ac     \/L2-R2' 

Let  the  ratio  of  connecting  rod  length  to  crank  length  be  denoted 
by  ft,  that  is 

L 


then  from  equation  (1) 


G 


(2) 


Let    p  =  the  allowable  bearing  pressure  in  pounds  per  square 

inch, 

a  =  length  of  bearing  surface  in  inches, 
6  =  breadth  of  bearing  surface  in  inches; 


PISTONS,   CROSS-HEADS  AND  STUFFING  BOXES         197 


then 


=  abp. 


(3) 


The   bearing   pressure  p  ranges   from  25  to   60   Ibs.,  the   lower 
values  being  used  for  engines  having  a  high  piston  velocity. 

EXAMPLE.  What  should  be  the  area  of  the  cross-head  slipper 
for  an  engine  using  steam  at  125  Ibs.  per  square  inch.  The  cylinder 
diameter  is  20  ins.  and  the  stroke  40  ins.  The  speed  is  102  R. P.M. 
The  connecting  rod  has  a  length  of  five  cranks.  Total  steam 
pressure  on  piston  is 

Q=^X202X  125  =  39,500; 


the  pressure  on  guides  is 


8000  Ibs. 


FIG.  19-14. 


.  19-15. 


The  mean  piston  speed  of  this  engine  is  680  ft.  per  minute.  This 
is  an  average  speed  for  medium-sized  engines  and  we  may  take 
the  allowable  bearing  pressure  at  40  Ibs.  per  square  inch,  then 

8000  =  40a&; 

and  the  bearing  area  of  each  slipper  is 

8000 


ab 


40 


200  sq.in.; 


or  length,  a,  may  be  18  ins.  and  breadth,  6,  about  11  ins. 

Wrist-pin.  This  is  designed  for  bearing  pressure  and  checked 
for  stress.  In  Fig.  19-15  are  shown  the  usual  types  of  wrist-  or 
cross-head  pins,  which  are  supported  at  both  ends.  They  have  a 
taper  fit  in  the  body  of  cross-head  and  are  further  prevented  from 


198 


ELEMENTS  OF   MACHINE   DESIGN 


rotating  by  a  pin  or  key.  In  a  the  pin  is  pulled  in  and  fastened 
by  means  of  a  nut  at  small  end  of  taper.  In  b  a  plate  at  large  end 
of  taper  presses  and  holds  pin  in  position.  The  hole  through 
center  of  pin  is  for  applying  the  lubricant  to  the  bearing  surface. 
The  allowable  bearing  pressure  ranges  from  700  to  1400  Ibs. 
per  square  inch.  The  method  of  calculating  the  'diameter  of  cross- 
head  pin  is  identical  with  that  of  crank-pin  in  Chapter  XVII. 
Using  the  same  symbols  we  obtain  the  same  equation 


:=  £ 

\nq 


(4) 


The  value  of  n  for  cross-head  pins  is  1.15  to  1.75.     Having  deter- 
mined diameter  and  length  according  to  equation  (4)  it  is  neces- 


FIG.  19-16. 


sary  to  check  design  for  stress.  Assuming  the  load  to  be  uniformly 
distributed  over  pin  (Fig.  19-16)  the  maximum  bending  moment 
is 

TUf      Pl 

^=-5-; 


and  therefore 


then 


pl 

- 


- 


Pl 


-^ 

rd? 


(5) 


The  stress  s  should  not  exceed  5000  Ibs.  per  square  inch. 


PISTONS,  CROSS-HEADS  AND  STUFFING  BOXES        199 

EXAMPLE.  Design  the  wrist-pin  for  the  engine  of  previous 
example. 

Total  thrust  on  wrist-pin  is 

P  =  39500-^ —     (equation  1,  Chapter  XVIII) 

-  40030  Ibs. 
Assume          q  =  800  and  n  =  1 .3 ; 

then 

/  40300  . 

and 

I  —  1 .3d  =  8£  ins.  (nearly) ; 

checking  for  stress,  we  have 

(_  40300X8.25 
S  (6.25)3 

which  is  within  safe  limits. 

Stuffing  Boxes.  Where  a  rod  having  either  a  rotary  or  recip- 
rocating motion  passes  through  the  wall  of  a  vessel  contain- 
ing a  fluid  under  pressure,  provision  must  be  made  to  prevent 
leakage.  The  device  used  for  this  purpose  is  called  a  stuffing 
box.  Thus  the  piston  rod  and  valve  rod  of  a  steam  engine  are 
examples  of  reciprocating  parts,  while  shafts  of  steam  turbines 
and  centrifugal  pumps  are  examples  of  rotating  parts  requiring 
stuffing  boxes.  The  packing  used  is  generally  some  soft  fibrous 
material,  but  for  high  pressures  and  temperatures  the  so-called 
metallic  packings  are  preferred. 

In  Fig.  19-17  is  shown  a  common  form  of  stuffing  box  for 
rods  from  1J  ins.  diameter  and  up.  It  consists  of  the  stuffing 
box  or  casing  A  and  the  gland  B.  The  gland  compresses  the 
packing  in  box  and  thus  tightens  it  about  rod.  The  proportions 
may  be  as  follows:  if  d  is  diameter  of  rod,  then  internal  diameter 
of  box  is 

D  =  1.25d+}in. 

The  length  of  box  is  made  from 

h  =  d+I  in.  to  1.5d-f-l  in., 


200 


ELEMENTS  OF  MACHINE  DESIGN 


high  pressures  requiring  the  longer  boxes.  For  the  smaller  sizes 
of  rods,  say  up  to  3  or  4  ins.  the  oval  gland  with  two  bolts  is  used, 
while  large  stuffing  boxes  commonly  have  glands  with  circular 
flanges  requiring  3  or  4  bolts.  The  diameter  of  bolts  may  be 


FIG.  19-17. 


FIG.  19-17a. 


For  rods  less  than  2  ins.  diameter  the  screwed  stuffing  box 
shown  in  Fig.  19-18  is  used.  The  gland  is  forced  into  box  by 
means  of  a  cap  threaded  over  box.  The  proportions  may  be  about 
the  same  as  those  given  above. 


PISTONS,   CROSS-HEADS  AND 'STUFFING  BOXES         201 

Large  rods  generally  have  the  gland  lined  with  a  brass  bush- 
ing and  if  there  is  considerable  side  pressure  on  rod,  as,  for  instance, 


FIG.  19-18. 


due  to1  weight  of  a  heavy  piston,  the  bottom  of  casing  also  is 
bushed,  as  shown  in  Fig.  19-17. 

A  type  of  metallic  stuffing  box  is  shown  in  Fig.  19-19.     The 
packing  consists  of  two  sets  of  split  rings  having  a  wedge-shaped 


FIG.  19-19. 

section.  A  spring  acting  on  a  sleeve  presses  the  rings  together, 
thus  tightening  the  inner  set  around  rod.  It  will  be  noticed  that 
the  sleeve  carrying  the  rings  is  on  a  spherical  seat;  this  enables 


202  ELEMENTS  OF   MACHINE  DESIGN 

the  rings  to  adjust  themselves  to  the  piston  rod  even  if  cylinder 
and  rod  axes  are  not  in  alignment. 

PROBLEMS 

1.  The  cross-head  shoes  of  a  steam  engine  are  6£  by  10  ins.     The  engine 
has  a  cylinder  12  ins.  in  diameter  by  18-in.  stroke.     The  length  of  connecting 
rod  is  40 \  ins.     Determine  the  maximum   bearing  pressure  per  square  inch 
on  cross-head,  if  the  steam  pressure  is  150  Ibs.  per  square  inch. 

2.  Determine  the    dimensions  of  the  wrist-pin  in  problem  1.     Assume 
bearing  pressure  is  800  Ibs.  per  square  inch  and  n  =  1.3.     Check  design  for 
bending  stress. 

3.  The  piston  rod  of  engine  in  problem  1  is  2  ins.  in  diameter.     Make  a 
sketch  design  of  stuffing  box  for  same,  to  scale. 

4.  Draw  to  scale  a  sketch  design  of  piston  for  engine  of  problem  1,  similar 
to  Fig.  19-4. 

•1  5.  In  a  gas  engine  the  maximum  pressure  in  cylinder  is  375  Ibs.  per  square 
inch.  The  cylinder  diameter  is  lOf  ins.  The  piston-pin  is  3£  ins.  in  diameter 
by  5|  ins.  long.  Determine  bearing  pressure  and  check-pin  for  stress. 


CHAPTER  XX 
HOISTING  MACHINERY  DETAILS 

Chains.  Link  chains  for  hoisting  machines  are  generally 
forged  to  gauge,  the  material  used  being  wrought  or  mild  steel. 
In  Fig.  20-1  are  shown  good  proportions  for  hoisting  chains.  The 
short  pitch  link  is  better  as  the  chain  is  more  flexible  and  when 
wound  on  drum  or  sheaves  is  subjected  to  smaller  bending  stresses. 


FIG.  20-1. 

The  long  pitch  chain  is  lighter  and  cheaper  for  the  same  strength. 
The  allowable  tensile  stress  in  chain  links  may  be  5000  Ibs. 
per  square  inch  for  power-driven  hoists  to  8000  Ibs.  for  hand- 
operated  machines.     This  gives  approximately 

P=12,000d2  (hand  driven), 
P  =  8000^  (power  driven), 

where  d  is  diameter  of  bar  from  which  chain  is  forged  arid  P  is 
safe  load. 

Sheaves  for  Chains.  In  Fig.  20-2  are  shown  various  forms  of 
sheaves  for  guiding  chains.  The  pitch  diameter  should  not  be 
less  than  20d.  The  sheave  shown  at  a  is  one  most  commonly 
used.  The  high  flanges  at  the  sides  are  necessary  only  where 

203 


204 


ELEMENTS  OF   MACHINE  DESIGN 


chain  is  subjected  to  much  vibration  or  where  it  is  led  off  at  an 
angle  to  the  plane  of  the  sheave. 

Chain  Drums.  These  are  provided  usually  with  a  helical 
groove  to  guide  the  chain  as 
in  Fig.  20-3.  A  clearance  of 
|  to  A  in.  is  allowed  between 
the  links  of  adjacent  turns. 
The  minimum  diameter  of 
drum  is  the  same  as  that  for 
sheaves.  The  minimum  thick- 
ness is  generally  determined 
by  the  possibility  of  obtaining 
a  sound  casting  and  is  not  less 


FIG.  20-2. 


FIG.  20-3. 


than  \  in.     The  following   empirical  equation   may  be  used  for 
obtaining  the  thickness  t  in  terms  of  drum  diameter  D. 


=  .02D+f  in. 


(D 


The  drum  may  be  checked  for  maximum  stress.     The  greatest 
bending  moment  occurs  when  load  acts  at  middle  of  drum  and  is 


M  = 


PL 
4  ; 


HOISTING   MACHINERY  DETAILS 


205 


the  twisting  moment  is 
TJ_ 

then  the  equivalent  bending  moment  is 


T     PD- 
T—~' 


and  the  stress  is 


PL        /PL      ,/PD 


Me                    Me 

Z          7T   /D4  —  Z>14\ 

32\      D 

) 

( 

(2) 
(3) 


FIG.  20-4. 

where  DI  is  the  internal  diameter  of  drum.  The  length  L  of 
drum  is  determined  by  the  maximum  distance  load  is  to  be  hoisted. 
When  the  load  is  in  its  lowest  position  two  or  three  turns  of  chain 
should  still  remain  on  drum. 

Chain  Wheels.  With  heavy  loads  and  high  lifts  chain  drums 
are  very  cumbersome  on  account  of  their  great  weight  and  large 
dimensions.  In  this  case  chain  wheels  are  used,  as  their  width 
is  independent  of  the  height  of  lift  and  their  diameter  may  be 
much  smaller  than  that  of  drums.  Such  a  wheel  is  shown  in  Fig 
20-5.  If  I  is  pitch  of  chain  and  N  is  number  of  teeth  on  wheel, 
then  the  pitch  diameter  of  wheel  is 


90°  1    '  I        90° 

sin  -i  F-  / .    \  cos 


(4) 


N  /  '     V~°  N  / 

For  wheels  having  eight  or  more  teeth  the  second  term  becomes 
negligible  and  we  may  write 

I 


D  = 


(5) 


206 


ELEMENTS  OF  MACHINE  DESIGN 


Hooks.     There  are  numerous  forms  of  hooks  used  as  experience 
developed  them  for  the  particular  use  to  which  they  are  to  be  put. 


FIG.  20-5. 


In  Fig.  20-6  and  20-7  are  shown  the  most  common  types.     The 
material    used  is  high-grade  wrought    iron  or  mild  steel.     The 


FIG.  20-6. 


FIG.  20-7. 


following  proportions  may  be  used,  referring  to  Fig.  20-8.     If  d\ 
be  the  outside  diameter  of  threaded  portion  of  hook,  then,  allow- 


HOISTING   MACHINERY  DETAILS 


207 


ing  a  stress  of  8000  Ibs.  per  square  inch  and  keeping  in  mind  that 
the  root  diameter  is  about  .83di,  we  have 


or  approximately 


(6) 


FIG.  20-8. 

then  the  other  dimensions  expressed  in  terms  of  d\  may  be 
<fc  =  4*,  ^3  =  ^1,  &i  =  di,  &2  =  £&i,  h  =  2di,  W=lMi  and  a  =  2#i. 
The  maximum  stress  occurs  in  the  section  MN.  This  section  is 
subjected  to  a  direct  tension,  st,  due  to  the  load  and  a  bending 
stress,  «&,  due  to  the  moment  arm  of  the  load  about  the  center 
of  gravity  of  the  section.  Then 


St=A> 


208 


ELEMENTS  OF  MACHINE  DESIGN 


where  A  is  the  area  of  the  section  and 

.     V** 

M    Px 

S6=7=T; 

where  x  is  the  distance  from  center  of  book  to  neutral  axis  of  sec- 
tion MN  and  the  total  stress  is 


(7) 


For  the  section  shown  in  Fig.  20-8 


The  proportions  shown  above  will  give  a  stress  of  about  18,000 
Ibs.  per  square  inch  at  the  section  MN.  This  is  allowable  if 
first-class  material  be  used. 


AW 


FIG.  20-9. 

Brakes.  The  simplest  form  of  brake  is  the  block  brake,  as  in 
Fig.  20-9.  The  friction  of  the  wood  block  against  the  brake  drum 
prevents  its  rotation. 

Let    W  =  weight  on  lever, 

Pt  =  force  to  be  braked  at  rim  of  drum, 
P  =  pressure  of  block  against  drum, 
M  =  coefficient  of  friction. 

Since,  in  order  to  have  equilibrium,  the  moment  of  all  'forces 
about  the  fulcrum  must  be  equal  to  zero,  we  have 


HOISTING  MACHINERY  DETAILS  209 

when  direction  of  rotation  is  as  indicated,  if  in  opposite  direction 


but  the  friction  at  rim  of  drum  must  at  least  be  equal  to  Pt  or 
then 


and  therefore 


(6) 


This  equation  shows  that  the  weight  W  necessary  to  hold 
load  in  equilibrium  is  proportional  to  Pt  the  tangential  force  at 
rim  of  brake  drum.  It  is  therefore 
desirable  to  place  brake  on  a  high- 
speed shaft  if  possible,  since  that 
gives  the  smallest  value  of  Pt.  If 
c  is  made  zero  the  last  term  in  equa- 
tion (8)  disappears.  Thus  when 
brake  drum  rotates  in  either  direc- 
tion by  making  c  equal  to  zero, 
the  weight  W  required  will  be  the 
same  for  both  directions  of  rota- 
tion. The  coefficient  of  friction  may 
be  taken  at  .4  for  wood  or  leather 
on  cast  iron,  if  surfaces  are  dry, 
and  .2  for  iron  on  iron.  To  avoid 
one-sided  pressure  on  the  brake 

drum  two  blocks  may  be  arranged  on  diametrically  opposite  sides 
of  drum,  as  in  Fig.  20-10.  By  the  use  of  grooved  friction  sur- 
faces the  braking  effect  may  be  greatly  increased  as  the  coeffi- 
cient of  friction  /*  in  equation  (8)  is  replaced  by 


FIG.  20-10. 


sn 


cos  a 


Band  Brakes.  In  this  type  of  brake  the  drum  is  encircled 
by  a  metallic  band  which  sometimes  is  lined  with  wood  or  other 
friction  surfaces,  This  band  may  be  tightened  around  drum  by 


210 


ELEMENTS  OF   MACHINE  DESIGN 


means  of  levers.  Fig.  20-11  shows  the  simplest  form  of  band 
brake.  One  end  of  band  is  attached  to  fulcrum  of  lever  while 
the  other  end  is  fastened  to  a  pin  a  short  distance  out  on  lever. 
It  is  evident  that  the  friction  of  band  on  drum  is  identical  to  that 


FIG.  20-11. 

of  a  belt  on  pulley  and  mathematically  it  may  be  treated  in  the 
same  way. 

Let      Ti  =  tension  on  tight  side  of  band, 
T2  =  tension  on  slack  side  of  band, 
^2  =  lever  arm  of  tension  T%, 
0  =  angle  subtended  by  band  in  radian  measure, 
/z  =  coefficient  of  friction, 
e  =  base  of  Napierian  logarithms  (2.718), 
P  =  force  or  load  at  end  of  lever, 
Pi  —  force  to  be  braked  at  brake  drum; 

then  (see  page  Appendix  B) 


HOISTING  MACHINERY  DETAILS 


211 


and  taking  moments  about  fulcrum  of  lever 
1         Pa-  7^2  =  0; 

•      P=T2l2=    Pt     l* 
"  a       e»'-l  a 


(9) 


In  this  equation  P  is  the  load  which  will  just  hold  a  force  Pt 
acting  at  the  rim  of  brake  drum,  in  equilibrium.  If  the  direction 
of  rotation  be  reversed  P  will  be  larger  since  now  the  greater 
tension  T\  has  a  lever  arm  h  about  the  fulcrum.  Proceeding  as 
above,  we  obtain  for  this  case  the  equation 


P  = 


e»e—l  a 


(10) 


The  value  of  /i  may  be  taken  at  .18  for  a  steel  band.     This 
gives  the  following  values  of  e»e: 

PORTION  OF  CIRCUMFERENCE  ENCLOSED  BY  BAND 


€*»  = 

.1 

.2 

.3 

.4 

.5 

.6 

.7 

.8 

.9 

1.12 

1.25 

1.40 

1.57 

1.76 

1.97 

2.21 

2.47 

2.77 

The  brake  band  is  designed  for  tension  with  an  allowable  stress 
of  8000  to  10,000  Ibs.  per  square  inch.  The  thickness  should  not 
exceed  a  maximum  of  ^  in.  while  the  breadth  should  not  be  greater 
than  3  or  4  ins.,  as  it  is  difficult  to  make  a  wide  band  lie  in  con- 
tact along  its  entire  width.  The  diameter  of  drum  is  usually 
from  8  to  16  ins.  unless  brake  is  part  of  hoisting  drum. 

The  differential  band  brake  is  illustrated  in  Fig.  20-12.  Here 
both  ends  of  band  are  attached  to  pins  located  at  different  dis- 
tances from  the  fulcrum.  A  counterweight  is  shown  to  balance 
the  weight  of  the  lever.  Taking  moments  about  fulcrum,  we  have 


and  therefore 


a 


substituting  for  T\  and  Tz  we  obtain 


(ID 


(12) 


212 


ELEMENTS  £>F   MACHINE   DESIGN 


EXAMPLE.  In  Fig.  20-13  is  shown  in  diagrammatic  form  a  hoist 
to  lift  a  load  of  5000  Ibs.  Find  the  force  P  required  to  hold  the 
load  in  equilibrium. 

The  force  Pt  which  must  be  applied  at  rim  of  brake  drum  is 


Pt 


X  5000  =  1900  Ibs. 


FIG.  20-12. 


FIG.  20-13. 

Assume  that  the  band  subtends  .7  of  the  drum  circumference, 
then 

^l  =  eM*=2.2     (from  table); 


HOISTING  MACHINERY  DETAILS 
and  from  equation  (9) 


213 


Fio.  20-14. 


FIG.  20-15. 


FIG.  20-16. 
The  maxium  tension  in  band  is 


214 


ELEMENTS  OF   MACHINE  DESIGN 


Ratchets  and  pawls  are  used  in  hoisting  machinery  to  prevent 
a  load  from  descending  when  the  operating  power  is  cut  off.  The 
ratchet  wheel  may  have  external  teeth  as  in  Fig.  20-14  or  internal 
as  in  Fig.  20-15.  With  external  teeth  the  angle  of  the  tooth  face 
should  be  such  that  when  the  pawl  is  in  engagement  the  common 
normal  will  pass  between  the  axes  of  pawl  and  ratchet;  for 
internal  teeth  the  normal  must  lie  outside  these  axes  as  shown  in 
figures.  The  teeth  are  calculated  for  bending.  Assuming  that 


FIG.  20-17. 

the  force  P  acts  at  point  of  tooth,  if  b  is  breadth  of  wheel,  then, 
from  Fig.  20-16  the  bending  moment  is 


and 


QPh 

ba2' 


(13) 


The  friction  stop,  Fig.  20-17,  has  the  advantage  of  noise- 
less operation.  The  angle  y  must  be  such  that  tan  7  is  less 
than  the  coefficient  of  friction  M  of  the  two  surfaces  in  contact. 
For  cast  iron  on  cast  iron  ^  =  .15,  thus  tan  7<.15  and  therefore 


HOISTING   MACHINERY  DETAILS 


215 


7  <8°.     To  avoid  this  small  angle  the  wheel  may  be  made  grooved 
as  in  Fig.  20-18.     In  this  case  y  is  determined  by  the  equation 

tan 


FIG.  20-18. 


PROBLEMS 

1.  The  drum  of  a  hoist  is  20  ins.  in  diameter  by  44  ins.  long.     The  thick- 
ness of  metal  is  |  in.     Determine  maximum  stress  in  drum  when  lifting 
30,000  Ibs. 

2.  Make  a  sketch  design  of  a  hook  according  to  proportions  given  in  Fig. 
20-8.     The  hook  to  carry  a  load  of  10  tons.     Determine  the  stress  in  sec- 
tion MN. 

3.  In  a  power-driven  hoist  the  drum  shaft  makes  1  revolution  to  5  of  the 
driving  shaft.     The  hoisting  drum  is  18  ins.  in  diameter,  the  brake  drum 
which  is  on  driving  shaft  is  15  ins.  in  diameter.    The  load  is  12,000  Ibs.     Design 
a  block  brake  similar  to  Fig.  20-9,  assuming  /z  =  .25.     Design  to  be  such  that 
one  man  can  operate  brake  for  either  direction  of  rotation.     Make  sketch 
of  your  design. 

4.  In  a  band  brake  similar  to  Fig.  20-11,  the  diameter  of  brake  drum  is 
30  ins.,  a  =  36  ins.,  £2=  4  ins.,  0  =  f  of  circumference  and  /*  =  .22.     Determine 
the  load  which  can  be  braked  at  the  brake  drum  if  force  at  end  of  lever  is 
60  Ibs. 


CHAPTER  XXI 
SPRINGS 

UNLIKE  other  machine  parts  springs  are  designed  to  give 
large  deflections  under  the  action  of  external  forces.  They  are 
used,  first:  to  maintain  definite  relative  positions  between  two 
machine  members  until  the  forces  acting  to  change  this  relative 
position  exceed  a  desired  limit,  as  in  valves,  governors,  etc.; 
second,  for  absorbing  energy  due  to  sudden  applications  of  force 
as  in  automobiles,  railway  cars,  etc.;  third,  to  store  energy  as 
in  clock  springs,  and  fourth,  to  measure  forces  as  in  spring  balances. 

Springs  may  be  divided  into  two  general  classes:  those  in 
which  the  chief  stress  is  a  bending  stress  and  those  in  which  the 
chief  stress  is  torsional.  In  the  following  discussion  let 

P  =  safe  load, 

/=  deflection  due  to  load  P, 
E  —  modulus  of  elasticity, 
Es=  modulus  of  torsional  elasticity. 

Flat  Spring.  (Fig.  21-1.)  This  is  the  simplest  form  of  spring. 
It  is  considered  to  be  a  cantilever  loaded  at  one  end,  then 


and 

72  « 

HOT  ........  (2) 

Another  form  of  this  spring  is  of  triangular  shape.     (Fig.  21-2.) 
The  safe  load  is  the  same  as  above.     The  deflection  is 


216 


SPRINGS 


217 


Laminated  or  Leaf  Springs.     These  consist  of  a  number  of 
layers  held  together  by  a  clip  as  shown  in  Fig.  21-3.     They  may 


FIG.  21-1. 


FIG.  21-2. 

be  considered  equivalent  to  the  lozenge-shaped  plate.  Due  to 
the  friction  between  leaves  the  deflection  will  be  somewhat 
less  than  calculated.  Such  springs  are  largely  used  in  motor  and 


218 


ELEMENTS  OF   MACHINE  DESIGN 


railway  cars,  where  they  are  generally  of  the  form  shown  in 
Fig.  21-4.     If  n  is  the  number  of  leaves 

then 

2  s6^  x^ 

^"3     I    n' 
and 

3  PI3 


J  —  Q 


8  bh3nE' 


(5) 


FIG.  21-3. 

Spiral  Springs.    (Fig.  21-5.)    These  springs  usually  have  a 
rectangular  section.    The  safe  load  is 

bh2s 


P  = 


6r  ' 


The  deflection  is 


2rls 


(6) 


(7) 


Helical  or  Coil  Springs.  (Fig.  21-6.)  In  this  type  of  spring 
the  chief  stress  is  torsion.  For  circular  section  of  wire  the  safe 
load  is 


SD  ' 


(8) 


SPRINGS 


219 


and  the  deflection  of  one  coil  under  a  load  P  is 

8D3P 


/= 


(9) 


In  the  design  of  a  coil  spring  the  maximum  load  P  and  the 
total  deflection  /'  required  is  known;  it  then  remains  to  find 
the  diameter  D  of  the  coil,  the  diameter  d  of  the  wire,  and  the 
number  of  coils  n.  As  we  have  only  two  equations  connecting 


FIG.  21-4. 


FIG.  21-5. 


FIG.  21-6. 


the  last  three  of  these  quantities  it  is  evident  that  we  may  assume 
any  one  of  the  three  and  calculate  the  other  two.  It  is  thus  possi- 
ble to  have  a  great  variety  of  springs  to  satisfy  the  same  conditions 
as  to  load  and  deflection.  To  facilitate  getting  practical  results 


the  following  tables  have  been  calculated, 
then  equation  (8)  becomes 


Let  the  ratio  -r  = 
a 


220 


ELEMENTS  OF  MACHINE  DESIGN 


In  Table  1  values  of  P  have  been  calculated  for  values  of  k  from 
4  to  20.  This  will  include  all  customary  sizes  of  springs.  It  has 
been  assumed  that  ss=  100,000,  a  value  which  may  be  used 
for  modern  high-grade  alloy  spring  steels.  For  any  other  value 
of  ss  it  is  simply  necessary  -to  multiply  the  tabular  value  by  the 
desired  ss  divided  by  100,000. 

•  The  deflection  has  been  tabulated  in  the  same  way  in  Table  2. 
If  in  equation  (9)  we  substitute  its  value  from  equation  (8)  and 

then  substitute  k  for  —  we  obtain: 
d 


Es   ' 

For  ss  =  100,000  and  Es  =  12,500,000  this  equation  becomes 
f=.Q251k2d.  The  deflections  given  in  table  are  based  on  these 
values. 

For  rectangular  sections  of  wire  (Fig.  21-7)  the  safe  load  is 


9    D  ' 


(10) 


and  the  deflection  per  coil  is 


FIG.  21-7. 


FIG.  21-8. 


Volute  or  Conical  Springs.  (Fig.  21-8.)  These  are  springs 
in  which  the  diameters  of  the  successive  coils  are  uniformly  re- 
duced. Since  the  maximum  stress  is  in  the  largest  coil  the  safe 
loads  are  the  same  as  those  of  coil  spring  of  the  same  section  and 
equal  diameter. 

Material  of  Springs.  High-carbon  steel  is  the  material  most 
commonly  employed  in  springs.  This  is  a  grade  containing  from 
.8  to  1.2  per  cent  of  carbon,  In  the  smaller  diameters  of  wire 


SPRINGS 


221 


i—  i  i—  i  <N  CO  -^  !>.  i—  I 


COi—  tOrH(NiO 
i—  i  05  CO  •*  10  00 


(N 


i—  i  C<1  CO  -^  CO  O5  CO 


COCS—iOOCOOiOOOO 


OiOOOO 
cOiOcOC^i—  i 


OOr^O^ 

i-HCOOO 


00 


CS    O^   Is* 


i—  iC<IOOOO 


|>  CO  >O  »O 


i—  ii—  iCO«OO5Tfri 


TfriOOTt< 

1-H    1-H    C<l 


CO 


>OCOl>-QOO5 


SI 

PQO 


222 


ELEMENTS  OF   MACHINE   DESIGN 


O 

<N 


O  iO  T—  i  O  O  O 


1—  IT-  lT-H<NCOCO^lOCOl> 


COCDCOC^OC^OOOOCOCOCDCOcOOOcO 

CDOOOCO»OO5<MCD'—  iCOCvQOOiOCOC^COCOeO-'fCOcOCOOOO 
i—  ii—  i(N<NC^<NCOCO<^<<t(tOiOcOI>OOO*OO»OOiOOO'—  i 


i—  i  i—  i  (N  <N  co  co  ->*i  >o  co 


oo  *o  10  co  to  i—  i  os  i—  i 

<NTjHcDOOOCOiOOi 


i— ixOO5^tlO'-HCsQCO'**l'O*OcOOOO 
COCOCOr^tOCOOilMiOOOr- ir^OCO 

'  ,-J  t-4  i-J  «*  oi  oo  co 


t^oooiOi-H 

OOOi-Hi— I 


!>•  t^*  I™H 

CNCOiO 


Tt<  t—  CO  iO  CO 


»O»-Ht^iOCO(N(N 


•<tf^cO<MT-Ht^tOtO 
CO  00  O  i 


00 


CO  CO  ^*  ^*  *O  *O  CO  !>•  OO  O^  ^^  ^H  CO  "^  CO  ^2  ^^  ^^  ^D  ^2  ^^  ^^  ^^  ^O 

OOOOOOOOOO'-H'-t'-lrHrH(NCOTjHlOCOt^OOO(M 


OOOOOOOOOOOOO^H'-ii-H 


•^i  iO  O^  00  l>*  >O  "^  00  O^  00 


0000 


CO    CO    T^    TjH 

0000 


CO  !>•  i— i  Oi  t>- 


is 


BB3Z 


O  O  O  O  O  O  ^H 


PQO 


SPRINGS  223 

or  rod  it  will  have  an  ultimate  tensile  strength  of  about  150,000 
Ibs.  per  square  inch,  after  proper  heat  treatment;  its  elastic  limit 
being  then  around  100,000  Ibs.  per  square  inch.  The  demands 
of  the  motor-car  industry  have  developed  a  number  of  special 
alloy  spring  steels  which  have  tensile  strengths  ranging  from 
200,000  to  300,000  Ibs.  per  square  inch  and  correspondingly 
high  elastic  limits.  Brass  and  bronze  is  also  used  extensively 
for  springs,  especially  in  the  small  sizes.  Besides  these  metals 
wood  and  rubber  is  sometimes  used  for  the  same  purpose. 

PROBLEMS 

1.  Design  a  helical  spring  for  a  safety  valve  so  that  it  will  open  when 
pressure  under  valve  is  150  Ibs.  per  square  inch.     The  valve  is  3  ins.  in 
diameter.      The  spring  is  to  be  about  5  ins.  long  when  compressed. 

2.  Design  the  helical  spring  for  the  exhaust  valve  of  a  50  H.P.  gas  engine. 
The  following  conditions  to  be  observed:     Force  on  spring  when  valve  is 
closed  200  Ibs.,  lift  of  valve  1J  ins.     Number  of  coils  15.     What  is  the  load 
of  spring  when  valve  is  open? 

/  3.  A  flat  spring  6  ins.  long,  1  in.  wide  and  ^  in.  thick  has  a  load  of  12  Ibs. 
acting  at  its  end.  Determine  deflection  of  spring.  (#  =  30,000,000.) 


CHAPTER  XXII 
MATERIALS  OF  MACHINERY 

THE  design  of  machinery  includes  a  great  deal  more  than  the 
knowledge  of  calculating  the  various  component  parts,  a  thorough 
acquaintance  with  the  materials  used  in  their  construction  is 
quite  essential  as  well  as  the  shop  methods  used  in  giving  the 
parts  their  required  shape.  In  the  following  pages  a  brief  outline 
of  the  most  commonly  used  materials  is  given. 

Iron.  By  far  the  most  important  material  which  enters  into 
the  construction  of  machines  is  iron.  It  is  used  in  any  of  three 
forms,  viz. :  cast  iron,  wrought  iron  and  steel.  The  raw  material 
for  all  these  is  iron  ore.  The  most  important  ores  are  magnetite 
or  black  oxide  of  iron,  which  when  pure  contains  72.4  per  cent  of 
iron.  The  famous  "  Swedish  iron "  is  made  from  this  ore. 
Haemetite,  or  red  oxide  of  iron,  which  is  the  chief  source  of  supply 
in  this  country,  contains  about  68  per  cent  of  iron.  Siderite,  or 
ferrous  carbonate,  which  is  an  important  source  of  supply  for  the 
English  iron  industry,  contains  about  48  per  cent  of  iron. 

The  smelting  process  takes  place  in  a  high  furnace  called 
blast  furnace.  The  charge  consists  of  the  ore,  fuel  and  a  flux. 
The  fuel  may  be  coke,  charcoal,  or  coal  which  is  free  from  sulphur, 
the  flux  being  limestone.  The  proportions  of  each  depend  upon 
the  ore  used.  The  materials  are  charged  at  the  top  of  the  fur- 
nace, which  is  50  to  100  ft.  high,  and  maintained  at  a  constant 
height  by  adding  new  material  every  ten  to  twenty  minutes  as 
the  melting  proceeds.  Near  the  base  of  the  furnace  a  blast  of 
air  is  admitted,  which  in  the  usual  hot  blast  method  has  been 
preheated  to  a  temperature  varying  from  500  to  1200°  F.  The 
pressure  of  the  blast  ranges  from  3J  to  10  Ibs.  per  square  inch. 
When  the  hearth  is  filled  with  the  molten  metal  the  blast  is  shut 
off  and  the  furnace  is  tapped. 

The  product  of  the  blast  furnace  is  known  as  pig  iron.  This 
is  classed  as  grey,  mottled,  or  white,  according  to  the  appearance 

224 


MATERIALS  OF   MACHINERY  225 

of  the  fracture.  Grey  pig  iron  has  a  granular  appearance,  a  dark 
grey  color  and  is  soft,  easily  machined  and  filed.  It  is  especially 
useful  for  foundry  purposes,  being  classified  as  Nos.  1,  2,  3, 
etc.,  according  to  the  greyness.  Mottled  iron  has  the  appearance 
of  a  matrix  of  white  with  grey  spots,  white  cast  iron  has  a  white 
and  often  crystalline  appearance.  It  is  extremely  hard  and  brittle, 
melts  more  readily,  but  flows  less  freely  than  grey  iron,  giving 
off  sparks  in  abundance.  It  is  less  suitable  for  making  castings, 
but  is  used  largely  in  the  production  of  wrought  iron. 

Pig  iron  contains  considerable  quantities  of  other  elements 
besides  iron;  it  is  these  impurities  which  largely  determine  the 
value  of  the  iron  for  different  purposes.  The  most  important  of 
these  substances  are  carbon,  silicon,  manganese,  sulphur  and  phos- 
phorus. The  carbon  exists  in  two  forms,  as  combined  carbon 
and  as  graphitic  carbon.  The  total  carbon  in  pig  iron  is  from  2  to 
5  per  cent.  The  greater  the  proportion  of  combined  carbon  the 
harder  and  whiter  the  resulting  product.  Graphitic  carbon  makes 
the  iron  grey  and  soft  and  raises  the  melting  point.  Sulphur  is 
looked  upon  as  the  chief  enemy  of  the  iron  and  steel  maker.  Its 
presence  in  wrought  iron  or  steel  causes  red  shortness,  that  is,  the 
metal  cannot  be  worked  well  above  a  red  heat,  but  cracks  under 
the  hammer. 

Cast  Iron.  This  is  the  product  of  the  blast  furnace  after  it 
has  been  remelted  in  the  cupola  of  the  foundry  and  is  formed  into 
the  desired  shape  by  pouring  into  a  mold.  A  good  deal  of  experi- 
ence and  judgment  is  required  by  the  foundryman  in  mixing  the 
various  grades  of  pig  iron  to  give  the  resulting  product  the  desired 
qualities.  Thus  a  steam  engine  cylinder  must  be  of  a  close-grained 
grey  cast  iron  soft  enough  to  be  easily  machined  but  not  so  soft 
as  to  wear  readily  due  to  the  rubbing  of  the  piston.  An  orna- 
mental casting  for  architectural  purposes  must  be  made  of  a  pig 
iron  containing  more  phosphorus  as  this  makes  it  expand  slightly 
on  solidifying  and  thus  gives  a  casting  true  to  the  mold. 

No  metal  used  by  the  machine  constructor  varies  so  much  in 
strength  and  soundness  as  cast  iron.  This  is  particularly  true 
of  its  resistance  to  tension,  which  varies  from  about  14,000  to 
about  30,000  Ibs.  per  square  inch.  Its  elastic  limit  is  low,  ordi- 
narily about  one-third  of  its  ultimate  resistance.  For  these  rea- 
sons a  high  factor  of  safety  should  be  employed,  for  dead  tensile 


226  ELEMENTS  OF   MACHINE  DESIGN 

loads  the  stress  allowed  may  be  taken  at  4000  to  5000  Ibs.  per 
square  inch.  The  compressive  strength  of  cast  iron  is  much 
higher,  ranging  generally  from  80,000  to  100,000  Ibs.  per  square 
inch. 

Malleable  Cast  Iron.  Malleable  castings  are  produced  by 
partially  decarbonizing  cast  iron.  This  is  done  by  placing  the 
castings  in  cast-iron  boxes  and  covering  them  with  an  oxidizing 
agent  such  as  haemetite  ore.  They  are  heated  and  maintained 
at  a  red  heat,  for  a  numbe*  of  days,  the  air  being  entirely  excluded. 
The  cast  iron  used  for  this  purpose  is  high  in  combined  carbon, 
viz.:  white  cast  iron;  a  portion  of  this  carbon  is  converted  by  this 
process  into  graphite  carbon.  In  this  country  the  so-called 
"  black  heart  "  process  is  used  in  which  the  casting  has  a  tough 
skin  resembling  wrought  iron  in  its  properties  and  a  soft  dark- 
grey  interior.  Such  castings  are  used  where  something  better 
than  cast  iron  is  wanted;  these  castings  approach  the  strength 
of  steel,  but  are  more  easily  machined.  Malleable  iron  should 
carry  a  central  load  of  3000  Ibs.  in  a  transverse  test  of  a  1-in. 
square  bar  on  supports  12  ins.  apart  with  a  deflection  of  j  in. 

Chilled  Castings.  The  greater  the  amount  of  combined  car- 
bon in  cast  iron  the  harder  the  resulting  casting.  Sudden  cooling 
tends  to  prevent  the  formation  of  free  or  graphitic  carbon.  This 
fact  is  made  use  of  in  the  production  of  chilled  castings.  In  this 
process  the  mold  or  a  portion  of  it  is  of  metal  with  a  thin  coating 
of  loam;  the  metal  is  rapidly  cooled  by  this  metal  wall  and  the 
resulting  casting  has  a  very  hard  surface,  the  effect  of  the  chill 
extending  a  distance  of  J  to  J  in.  below  the  surface.  Such  cast- 
ings are  used  where  great  hardness  is  needed  to  resist  wear  and  a 
strong  tough  interior  to  withstand  shock.  Thus  car  wheels  and 
the  rolls  used  in  rolling  mills  are  often  chilled  castings. 

Wrought  Iron.  This  is  the  purest  form  of  commercial  iron 
containing  in  the  best  grades  up  to  99.6  per  cent  iron  and  only 
0.4  per  cent  of  impurities,  chiefly  carbon  and  slag.  Wrought  iron 
is  made  by  oxidizing  the  impurities  of  pig  iron  in  a  reverberatory 
furnace.  This  method  is  called  the  puddling  process.  The  raw 
pig  iron  used  must  be  of  a  special  grade,  low  in  sulphur,  phos- 
phorus, and  silicon;  charcoal  pig  is  the  material  which  best 
fulfills  these  conditions. 

In  the  reverberatory  or  puddling  furnace  the  fuel,  unlike  the 


MATERIALS  OF  MACHINERY  227 

blast  furnace,  does  not  come  in  contact  with  the  charge  of  pig 
iron;  the  flames  and  hot  gases  of  combustion  heating  the  charge 
to  the  melting  point.  The  hearth  on  which  the  charge,  400  or 
500  Ibs.,  is  placed  is  lined  with  an  oxidizing  material,  called 
fetling,  such  as  iron  oxide  ore;  in  about  half  an  hour  it  is  partially 
melted,  forming  a  pasty  mass  which  becomes  less  and  less  fluid 
as  the  carbon  and  other  impurities  are  removed.  It  is  stirred 
with  an  iron  tool  called  a  rabble  so  as  to  bring  all  parts  under  the 
influence  of  the  fetling.  The  metal  is  then  collected  in  balls 
or  blooms  weighing  about  80  Ibs. 

It  is  now  in  a  soft  spongy  condition  quite  unfit  for  use.  These 
blooms  are  subjected  to  a  hammering  or  squeezing  process  which 
removes  a  portion  of  the  slag  present.  The  resulting  product  is 
known  as  puddled  bar  and  is  still  of  too  coarse  a  structure  to  be 
used.  The  puddled  bar  is  cut  up  into  short  lengths  which  are  piled 
into  a  faggot,  reheated  to  a  welding  heat  and  again  hammered  and 
rolled  into  bars,  which  are  commercially  known  as  merchant 
bar.  This  process  of  cutting,  piling,  reheating  and  rolling  is  re- 
peated a  number  of  times,  depending  on  the  quality  of  the  wrought 
iron  desired.  Each  repetition  makes  the  metal  purer  and  more 
uniform  in  structure,  giving  it  a  fibrous  appearance.  The  name 
of  the  resulting  product  gives  the  number  of  times  the  above  proc- 
ess has  been  repeated;  thus  best  bar  is  made  from  merchant 
bar,  best  best  from  best  bar  and  treble  best  from  best  best. 

Although  steel  has  largely  displaced  wrought  iron,  partly 
because  it  is  about  20  per  cent  cheaper,  there  are  many  engineers 
who  prefer  wrought  iron  for  specific  purposes,  as  for  rivets,  bolts, 
etc.  Good  wrought  iron  should  have  a  tensile  strength  of  45,000 
to  55,000  Ibs.  and  an  elongation  of  25  per  cent  before  rupture 
occurs,  and  have  an  elastic  limit  of  about  one-half  its  tensile 
strength,  and  it  should  bend  double,  cold,  without  cracking. 

Case  Hardening.  A  process  by  means  of  which  the  surface 
of  a  wrought-iron  article  is  made  to  absorb  carbon  and  is  thereby 
given  the  property  of  becoming  hard  upon  heating  and  suddenly 
quenching,  is  called  case  hardening.  The  method  is  as  follows: 
The  articles  to  be  case  hardened  are  packed  in  cast-  or  wrought- 
iron  boxes  together  with  some  powdered  material  rich  in  carbon 
such  as  bone  dust  or  animal  charcoal;  often  these  are  mixed  with 
some  cyanide.  Each  article  must  be  entirely  surrounded  and  in 


228  ELEMENTS  OF  MACHINE  DESIGN 

close  contact  with  the  carbonizer.  The  box  is  closed  by  a  cover 
which  is  sealed  down  air  tight  with  clay.  The  boxes  are  now 
placed  on  a  furnace  and  heated  to  a  good  orange  heat.  This 
temperature  is  maintained  from  two  to  twenty-four  hours,  depend- 
ing on  the  size  of  the  articles  and  depth  to  which  they  are  to  be 
case  hardened.  The  boxes  are  now  withdrawn  from  the  furnace 
and  allowed  to  cool  slowly.  The  articles  are  then  removed 
and  cleaned.  They  are  then  heated  in  a  muffle  furnace  to  a  cherry 
red  and  quenched  in  oil  or  water. 

Steel.  The  carbon  contents  of  iron  largely  determine  the 
difference  between  cast  iron,  wrought  iron  and  steel.  Thus 
wrought  iron  is  very  low  in  carbon  contents,  from  .10  to  .40  per 
cent;  cast  iron  is  rich  in  carbon,  containing  from  3  to  5  per  cent 
of  it.  Steel  is  intermediate  between  these,  containing  from  .15 
to  1.8  per  cent  carbon.  According  to  the  amount  of  carbon  it  is 
classified  into  low  carbon  or  mild  steel  and  high  carbon  or  tool 
steel. 

Mild  Steel.  The  raw  material  for  mild  steel  is  pig  iron.  There 
are  two  methods  of  manufacture,  the  Bessemer  process  and  the 
open-hearth  process.  Mild  steel  can  be  easily  forged,  rolled  and 
welded,  but  cannot  be  hardened  except  by  the  case  hardening 
process. 

The  Bessemer  process  of  making  steel  is  carried  on  in  a  pear- 
shaped  vessel  called  the  Bessemer  converter.  The  bottom  of  this 
vessel  is  perforated  with  many  small  holes  through  which  a  power- 
ful blast  of  air  enters.  The  converter  is  partly  filled  with  a  bath 
of  molten  pig  iron.  The  action  of  the  air  blast  is  to  burn  out  nearly 
all  the  silicon  manganese  and  carbon.  The  converter  is  lined  with 
either  an  acid  or  a  basic  lining  and  is  mounted  on  trunnions  so  that 
it  can  be  tilted  to  any  angle.  After  the  impurities  have  been 
burned  out  it  is  necessary  to  recarburize  the  bath.  This  is 
usually  done  by  adding  some  material  rich  in  carbon,  after 
the  converter  has  been  emptied  into  a  ladle.  The  time  required 
for  a  heat  is  from  eight  to  fifteen  minutes  in  the  acid  process  and 
from  twenty  to  twenty-five  minutes  in  the  basic  process,  the  ca- 
pacity of  the  converters  is  from  8  to  25  tons  per  heat. 

Open-hearth  Steel.  In  this  process  the  charge  consists  of 
either  pig  iron  or  of  pig  iron  and  scrap  and  ore.  This  is  melted 
in  the  hearth  of  a  regenerative  Siemens  furaace.  Here,  as  in  the 


MATERIALS  OF  MACHINERY  229 

Bessemer  process  there  is  a  basic  and  an  acid  process  according 
to  the  lining  being  acid,  silica  bricks  and  sand,  or  basic,  magnesite 
brick  covered  with  crushed  dolomite.  These  furnaces  are  made 
in  sizes  from  5  tons  up  to  200  or  over.  A  heat  takes  from  five 
to  ten  hours.  The  production  of  open-hearth  steel  is  rapidly 
increasing  and  it  seems  to  be  supplanting  Bessemer  steel  in  many 
fields.  This  reason  for  this  is  that  it  is  a  sounder  and  more 
homogeneous  metal. 

High-carbon  Steel.  This  steel  is  manufactured  either  by  the 
crucible  process  or  by  the  cementation  process;  the  raw  material 
being  either  wrought  iron,  or  mild  steel.  The  cementation  process 
is  used  chiefly  in  the  Sheffield  district  in  England.  In  this  process 
bars  of  pure  wrought  iron  are  impregnated  with  carbon.  For 
this  purpose  bars  2J  to  3  ins.  wide,  f  to  f  in.  thick  and  about  12  ft. 
long  are  packed  in  layers  separated  by  charcoal  in  a  cementing 
furnace.  The  furnace,  which  is  sealed  air-tight  and  heated  exter- 
nally, attains  its  full  temperature  in  three  or  four  days  and  is 
maintained  at  this  for  about  seven  to  twelve  days,  depending  on  the 
carbon  content  desired  in  the  steel.  The  charge  is  then  cooled, 
this  requiring  from  four  to  six  days.  The  resultant  product  is 
known  as  blister  steel.  It  is  of  a  coarse  crystalline  structure. 
The  blister  steel  is  broken  into  short  lengths,  piled  into  faggots, 
these  are  reheated  and  hammered  or  rolled  into  bars,  this  product 
being  known  as  shear  steel.  This  greatly  improves  the  quality 
of  the  steel,  and  the  process  may  be  repeated  several  times, 
depending  on  quality  of  product  desired. 

In  the  crucible  process  high-carbon  steel  is  produced  by  melting 
the  charge,  consisting  usually  of  Swedish  wrought  iron,  in  crucibles 
with  the  addition  of  some  carbon.  The  crucibles,  which  have  a 
capacity  of  about  100  Ibs.,  are  heated  in  a  gas  furnace  of  the 
regenerative  type.  It  requires  3^  to  4  hours  to  convert  a  charge 
into  steel.  High-carbon  steel  produced  by  the  crucible  process 
is  known  as  cast  steel.  This  as  well  as  the  high-carbon  steel 
produced  by  the  cementation  process  is  used  for  all  varieties  of 
cutting  tools  and  wherever  great  hardness  and  good  wearing 
qualities  are  required.  The  high-carbon  steels  possess  the 
property  of  hardening  when  heated  and  quenched. 

Alloy  Steels.  Alloy  steels  are  those  which  owe  their  properties 
chiefly  to  the  presence  «f  an  element,  or  elements  other  than  carbon. 


230  ELEMENTS  OF  MACHINE  DESIGN 

Nickel  Steel.  This  is  the  most  widely  used  of  all  alloy  steels. 
The  nickel  contents  vary  from  1.5  to  4.5  percent,  the  most  usual 
proportion  being  3  to  3.75  per  cent,  the  carbon  varying  from  .2  to 
.5  per  cent.  Nickel  steel  is  largely  used  for  engine  forgings  (con- 
necting rods,  etc.),  locomotive  axles,  shafting  and  motor  car 
frames  and  engine  parts.  Nickel  steel  has  a  higher  elastic  limit 
than  carbon  steel;  this  enables  it  to  resist  repeated  and  alternating 
stresses  better.  The  ultimate  tensile  strength  is  increased  by 
about  4000  Ibs.  per  square  inch  above  that  of  carbon  steel  by 
each  per  cent  of  nickel  added  up  to  5  per  cent.  It  is  somewhat 
harder  than  ordinary  steel  but  not  so  much  but  that  it  can  be 
readily  machined.  It  is  also  used  for  making  steel  castings,  as 
it  is  comparatively  free  from  blow  holes  and  melts  at  a  lower  tem- 
perature than  carbon  steel.  Nickel  steel  has  a  very  small  co- 
efficient of  expansion.  When  the  amount  of  nickel  is  36  per  cent  the 
coefficient  is  less  than  that  of  any  other  metal  or  alloy  known, 
being  practically  zero.  This  alloy  has  been  patented  and  goes 
under  the  trade  name  of  Invar.  Platinite  is  another  nickel  steel 
alloy  containing  42  per  cent  of  nickel  and  having  the  same  coef- 
ficient of  expansion  as  glass. 

Chrome  Steel.  This  alloy  when  in  the  hardened  state  com- 
bines the  properties  of  great  hardness  with  high  elastic  limit. 
The  chromium  ranges  from  1  to  2  per  cent  with  carbon  from  .8 
to  2  per  cent.  It  is  used  for  making  armor-piercing  projectiles, 
for  very  hard  plates  such  as  are  used  in  burglar-proof  safes,  and 
where  for  reasons  of  wear  great  hardness  is  required,  as  in  auto- 
mobile gears. 

Vanadium  Steel.  The  manufacture  of  this  the  newest  of  the 
alloyed  steels  is  still  in  its  infancy.  An  addition  of  only  .1  to 
.15  per  cent  of  vanadium  greatly  increases  the  strength  of  steel. 
Its  chief  characteristic  is  its  resistance  to  torsional  and  vibratory 
stresses.  It  is  therefore  useful  where  lightness  and  strength  are 
required,  as  in  springs  and  in  auto  axles,  etc. 

Tungsten  Steel.  This  steel  is  also  known  as  self -hardening, 
air-hardening  or  high-speed  steel.  The  amount  of  tungsten  used  is 
as  high  as  24  per  cent  but  usually  ranges  from  5  to  10  per  cent  with 
carbon  from  .4  to  2  per  cent.  These  steels  cannot  be  annealed 
by  any  known  process,  and  will  maintain  a  cutting  edge  even 
at  a  red  heat.  Messrs.  F.  W.  Taylor  and  M.  White  made  a  very 


MATERIALS  OF  MACHINERY  231 

extensive  set  of  tests  with  this  steel  for  the  Bethlehem  Steel 
Works.  The  results  of  these  tests  were  the  beginning  of  a  new 
era  in  machine-shop  practice. 

Non-ferrous  Metals.  Next  to  iron  the  most  important  metal 
that  is  used  by  the  machine  constructor  is  copper.  Its  great 
value  lies  in  the  ease  with  which  it  alloys  with  other  metals  to 
form  an  almost  infinite  number  of  very  useful  alloys.  Just  as 
different  bodies  are  soluble  to  varying  degrees  in  water  or  even 
entirely  insoluble,  so  metals  possess  the  property  of  alloying  with 
others  in  different  degrees,  some  perform  this  readily  in  all  propor- 
tions, as  for  instance  copper  with  tin,  copper  with  zinc,  iron  with 
manganese,  and  gold  with  silver;  others  only  in  a  limited  degree, 
as  iron  with  zinc,  or  iron  with  copper;  and  others  not  at  all,  as 
iron  with  lead  or  with  silver. 

The  purpose  of  alloying  metals  is  to  produce  a  material  pos- 
sessing certain  valuable  properties  not  possessed  by  the  individual 
constituents.  In  general  the  tenacity  is  increased  if  certain  ratios 
of  the  constituents  are  not  exceeded.  The  melting-point  is 
usually  lower  than  the  mean  of  the  elements  which  make  up  the 
alloy.  Thus  tin  melts  at  230°  C.,  and  lead  at  330°  C.,  while 
the  alloy  tin  63  parts  and  lead  37  parts  melts  at  181°  C.  Heat 
and  electric  conductivity  are  always  decreased. 

Brass.  Nominally  brass  should  consist  of  two  metals,  copper 
and  zinc  only,  but  many  varieties  contain  small  amounts  of  iron, 
tin,  arsenic  and  lead.  As  copper  and  zinc  will  alloy  in  almost  all 
proportions  a  great  variety  of  brass  alloys  exist.  The  most 
useful  may  be  divided  into  three  grades  of  soft,  medium,  and  hard 
brass  having  approximately  the  following  composition,  soft, 
copper  80  and  zinc  20;  medium,  copper  70  and  zinc  30;  and  hard, 
copper  60  and  zinc  40.  The  medium  is  used  for  sheets,  tubing 
and  wire.  It  is  readily  worked  cold  but  not  so  easily  hot.  The 
hard  is  well  adapted  for  casting,  as  it  gives  sharp  and  clean  cast- 
ings and  is  cheap  on  account  of  the  high  zinc  contents.  It  forges 
well  in  the  hot  state  and  is  therefore  adapted  for  drop  forgings. 
The  80  and  20  alloy  is  used  where  a  special  soft  material  is  desired. 

Brass  castings  are  soft  whether  suddenly  or  slowly  cooled. 
Rolling  or  otherwise  mechanically  working  brass  makes  it  hard 
and  brittle.  It  is  therefore  necessary  to  anneal  it,  if  further  reduc- 
tion is  required.  There  are  a  number  of  brass  alloys  which  have 


232  ELEMENTS  OF  MACHINE  DESIGN 

been  patented  or  are  known  by  special  trade  names.  Thus  Muntz 
metal  consists  of  copper  56  to  63  per  cent  and  zinc  44  to  37  per 
cent.  Delta  metal,  55  copper,  43J  zinc,  1  iron  and  small  quanti- 
ties of  lead  and  phosphorus. 

Bronze.  This  alloy,  which  has  been  in  the  service  of  man  for 
unknown  centuries,  is  primarily  a  combination  of  copper  and  tin. 
There  is  an  exceedingly  wide  range  in  which  these  two  metals  alloy 
and  a  corresponding  variation  in  physical  properties  of  the  result- 
ant product.  Standard  commercial  bronze  alloys  contain  tin 
and  copper  in  ratios  varying  from  1  tin  and  4  copper,  to  1  tin  and 
12  copper.  Thus  gun-metal  bronze  contains  copper  89  to  92 
per  cent  and  tin  11  to  8  per  cent;  bearing-bronze  copper  82  to 
88  per  cent  and  tin  18  to  12  per  cent. 

Phosphor  bronze  contains  from  J  to  1  per  cent  of  phorphorus. 
This  alloy  is  much  used  in  high-grade  construction  for  bearings, 
also  for  springs,  gear  blanks,  etc. 

Tobin  bronze  is  used  most  frequently  in  the  rolled  shape  for 
bolts,  pump  piston  rods,  etc.,  where  it  is  subject  to  corrosion  by 
salt  water.  As  used  in  the  U.  S.  navy  its  composition  is  copper 
59  per  cent,  zinc  38.4  per  cent,  tin  2.16  per  cent,  lead  .31  per  cent, 
and  iron  .11  per  cent. 

Manganese  bronze  is  an  alloy  of  copper  and  zinc  improved  by 
small  quantities  of  manganese,  iron,  phosphorus  and  tin.  It 
varies  considerably  in  composition  and  may  be  either  soft  and  duc- 
tile or  hard  and  strong.  It  is  almost  universally  used  for  the  pro- 
pellers of  large  steamboats,  as  it  is  strong,  resists  corrosion  by 
salt  water,  and  has  a  very  smooth  surface  when  cast.  Its  tensile 
resistance  is  from  65,000  to  75,000  Ibs.  per  square  inch.  There 
are  numerous  other  bronzes  each  possessing  some  desirable 
property,  as  for  example,  chromax  bronze  has  a  tensile  strength' 
equal  to  that  of  high-grade  steel,  viz.:  70,000  to  80,000  Ibs.  per 
square  inch.  It  is  also  an  alloy  of  copper,  zinc,  nickel,  aluminum 
and  chromium. 

White-metal  Alloys.  These  metals  are  the  so-called  anti- 
friction metals  used  for  lining  bearings.  There  is  an  exceedingly 
large  variety  on  the  market,  some  of  high-sounding  name  and 
inferior  quality.  They  may  be  divided  into  four  classes:  1st, 
tin,  antimony,  and  copper  alloys;  2d,  lead  and  antimony  alloys; 
3d,  lead,  tin  and  antimony  alloys;  and  4th,  zinc,  tin  and  anti- 


MATERIALS  OF  MACHINERY  233 

mony  alloys.  There  are,  however,  numerous  variations  possible; 
copper  sometimes  displaces  the  zinc  or  antimony  in  the  last  two 
classes. 

The  best  known  and  most  widely  used  of  these  white-metal 
alloys  are  the  Babbitt  metals.  The  original  Babbitt  metal  was  an 
alloy  containing  about  89  per  cent  tin,  7  to  8  per  cent  antimony 
and  3  to  4  per  cent  copper;  but  very  little  of  the  so-called  genuine 
Babbitt  metal  that  is  sold  is  made  according  to  the  original 
formula.  As  a  matter  of  fact  almost  any  white  metal  that 
is  used  for  lining  bearings  is  called  Babbitt  metal.  Tin  being 
by  far  the  most  expensive  of  the  metals  used  in  these  bearing 
alloys,  the  much  cheaper  lead  and  antimony  compounds  are 
preferred. 

These  white-metal  alloys  can  be  easily  melted  in  an  ordinary 
ladle.  It  is  therefore  a  simple  process  to  pour  the  alloy  around 
the  journal  into  the  bearing  body,  and  for  small  or  medium-sized 
bearings  this  is  usually  done.  For  low  or  medium  speeds  and  bear 
ing  pressures  that  are  not  too  high,  depending  on  the  hardness  of 
the  alloy,  these  white-metal  alloys  make  the  best  possible  bear- 
ing linings. 

Aluminum,  one  of  the  newest  metals  at  the  command  of  the 
engineer,  is  of  particular  advantage  where  great  lightness  is  desir- 
able, as  its  weight  is  only  about  one-third  tha*  of  iron.  It  is 
very  malleable  and  ductile;  it  can  be  cast  in  molds  similar  to  cast 
iron.  In  the  form  of  castings  it  has  a  tensile  resistance  of  15,000 
Ibs.  per  square  inch  which  can  be  increased  to  25,000  by  rolling 
or  hammering.  Combined  with  copper  it  forms  some  valuable 
alloys  known  as  aluminum  bronzes,  the  tensile  strength  of  which 
run  up  to  90,000  Ibs  per  square  inch. 

Timber.  The  term  timber  is  applied  both  to  standing  trees 
and  to  lumber  of  large  dimensions.  Lumber  is  a  general  term 
applied  to  wood  which  has  been  cut  into  any  of  the  various  sizes 
used  in  construction  work.  The  trees  from  which  our  lumber  is 
derived  may  be  divided  into  broad-leaf  or  hard  wood  and  needle 
leaf  or  soft  wood. 

If  a  section  of  a  tree  be  examined  it  will  be  seen  to  consist  of 
a  somewhat  darker  interior  core  called  heart  wood,  a  lighter 
exterior  portion  called  sap  wood  surrounding  this  bark.  The  heart 
wood  and  sap  wood  are  arranged  in  concentric  rings,  one  of  which 


234  ELEMENTS   OF   MACHINE  DESIGN 

is  formed  each  year.  Dense  narrow  rings  which  indicate  slow 
growth  produce  strong  wood. 

Green  wood  contains  a  good  deal  of  moisture,  the  effect  of  which 
is  materially  to  reduce  the  strength.  It  is  therefore  necessary  to 
dry  the  wood  before  using.  If  done  in  the  open  it  is  said  to  be  air 
dried,  a  method  which  may  take  from  several  months  to  two  or 
three  years.  If  it  is  dried  in  an  enclosed  chamber  to  which  heat 
is  applied  it  is  kiln-dried  lumber,  this  being  a  much  more  rapid 
method.  The  process  of  drying  wood  is  called  seasoning. 

The  strength  of  timber  depends  on  the  amount  of  heart  wood 
or  sap  wood  and  such  defects  as  knots,  checks,  cracks,  and  any 
break  in  the  continuity  of  fibers.  Wood  is  much  stronger  both 
in  tension  and  compression  along  the  grain  than  across  the  grain. 

A  brief  description  of  the  physical  properties  and  uses  of  the 
more  common  woods  is  given  below. 

White  Ash.  Heavy,  hard,  strong,  elastic,  becoming  brittle 
with  age.  Not  durable  in  contact  with  soil.  Uses:  Agricultural 
implements,  handles,  oars,  interior  and  cheap  cabinet  work. 
Red  and  green  ash  are  frequently  used  as  substitutes  for  the  more 
valuable  white  ash. 

White  Cedar.  Soft,  light,  fine  grained,  durable  in  contact  with 
soil,  but  not  very  strong  or  tough.  Uses :  cooperage,  boat  build- 
ing, shingles,  posts,  etc. 

Cypress.  Light,  hard,  close  grained  but  brittle.  An  exceed- 
ingly durable  wood,  easily  worked,  takes  a  high  polish  with  satiny 
appearance.  Uses:  interior  finish  and  cabinet  work,  construc- 
tion, etc. 

White  Elm.  Heavy,  hard,  strong  and  tough.  Not  easily 
shaped  and  warps  in  drying.  Takes  good  polish.  Uses:  car, 
wagon  and  ship  building,  bridge  timbers,  sills,  furniture,  and  bar- 
rel staves. 

Hickory.  Heaviest,  hardest  and  toughest  of  American  woods. 
Very  flexible.  Uses:  handles,  and  best  wood  implements. 

Hard  Maple.  Tough,  heavy,  hard  and  strong.  Takes  good 
polish.  Not  durable  when  exposed.  Uses:  furniture,  flooring, 
pegs,  interior  finish,  etc. 

Soft  Maple.  Light,  brittle,  and  easily  worked,  moderately 
strong.  Takes  high  polish.  Not  durable  when  exposed.  Uses: 
woodenware,  turned  work,  flooring,  interior  finish,  etc. 


MATERIALS  OF  MACHINERY  235 

White  Oak.  Heavy,  strong,  tough  and  close  grained.  Takes 
a  high  polish.  Uses:  framing  structures,  ship  building,  interior 
finish  and  furniture  making. 

White  Pine.  Light,  soft,  and  straight  grained.  Easily 
worked,  but  not  very  strong.  Uses:  interior  finish,  pattern 
making,  etc. 

Yellow  Pine  (long-leaf).  Heavy,  strong,  hard,  tough,  durable 
if  dry  and  well  ventilated,  but  not  in  contact  with  ground.  Uses: 
heavy  framing  timbers,  flooring,  etc.  Short-leaf  yellow  pine 
is  used  as  a  substitute  for  the  long  leaf. 

Oregon  Pine  (Douglas  Fir).  Hard,  strong  but  of  variable 
quality,  durable  but  difficult  to  work.  Used  for  all  kinds  of 
construction. 

Black  Spruce.  Light,  soft,  close  and  straight  grained.  Used 
for  piles,  framing  timbers,  submerged  cribs,  etc. 


APPENDIX  A 

If  in  Fig.  7-10  an  elementary  ring;  of  width  dr  be  taken  then 
the  moment  of  the  friction  on  this  ring  is  the  total  pressure  on  the 
surface  multiplied  by  the  coefficient  of  friction  and  by  the  radius 
n  of  the  ring,  or 


oooooo 


(1) 


but 
or 


substituting  this  value  in  equation  (2)  we  obtain 


If  the  discs  are  in  the  form  of  a  ring  of  outside  radius  R  and 
inside  radius  r,  then  equation  (1)  above  becomes 

*-r3)-,     ...     (3) 


but  Pa  =  Tr(R2  —  r2)  and  substituting  this  value  (3)  reduces  approx- 
imately to 

(4) 


2 


237 


APPENDIX  B 

Let  Fig.  9-2  represent  an  elementary  length  of  belt  on  the 
rim  of  its  pulley.  Relatively  to  the  pulley  it  is  in  equilibrium 
under  the  forces  shown,  that  is,  a  tension  T  at  one  end,  T-\-dT  at 
the  other  end,  and  friction  dF,  or 

T+dT=T+dF  .........     (1) 

If  dN  is  the  normal  pressure  between  belt  and  pulley,  then 


but  from  the  parallelogram  of  forces 


and  since  d6  is  an  infinitesimal  angle  we  can  substitute  for  the 
sine  the  angle  itself,  then 


Neglecting  the  product  of  the  infinitesimals, 

dN  =  Td6; 
therefore 

dF=nTd8; 
then  from  equation  (1) 

dT  =  dF; 


integrating  we  obtain 

CTl  rIT 

L  ?= 

or 


238 


APPENDIX  C 

In  Fig.  14-5  is  shown  a  portion  of  the  pipe.     The  total  radial 
pressure  on  an  arc  subtending  the  angle  d6  is 


=  rd6Lp', 

and  the  component  of  this  pressure  normal  to  plane  AB  is 
ON  =  MN  cos0 

=  rLp  cos  6dB. 
The  total  vertical  component  R  is  therefore 

+1 
rLp  cos  BdQ 

i 

2rLp  =  DLp. 


239 


INDEX 


PAGE 

Acme  thread 22 

Allowance  for  shrink  and  force  fits 56 

Aluminum 233 

Axles 62 

Beams,  table  of 11 

Bearing  pressure 75 

Bearings 78 

,  adjustment  of 80 

,  ball 86 

,  ball  thrust 88 

,  friction  of  thrust 84 

,  oil  grooves  for 80 

,  roller 91 

,  thrust 82 

Belt  fasteners 93 

Belts 93 

,  rules  for  installation 90 

Belt  transmission 94 

,  practical  methods  of  design 98 

Bending 6 

Bending  moment 8 

Boltheads,  forms  of 24 

Bolts,  multiple  threaded 26 

,  strength  of 26 

,  types  of 23 

Brakes,  band 209 

,  block 208 

Brass 231 

Bronze 232 

Buckling 12 

Butt  joint,  double  riveted 41 

,  single  riveted 41 

,  triple  riveted 42 

Buttress  thread 22 

Calking 34 

Case-hardening 227 

241 


242  INDEX 

PAGE 

Chain  drums 204 

Chain  gearing 134 

Chain  wheels 205 

Chains,  hoisting 203 

Chilled  castings 226 

Clutch,  coil 73 

,  cone 69 

,  cylindrical 72 

,  disc 71 

,  friction 68 

,  pin 68 

,  positive . . 67 

Cocks 150 

Columns,  table  of 13 

Connecting  rod  ends 179 

,  design  of 185 

Connecting  rods 179 

,  design  of 182 

Cotters 49 

,  strength  of 51 

Coupling,  flange 64 

,  flexible 66 

,  Sellers 65 

,  sleeve 64 

,  universal 66 

Cross-heads 194 

,  design  of 196 

Cranked  shafts 167 

,  calculation  of .  . ; 168 

Crank  pins '. 172 

Cranks 170 

Crank  shafts,  stresses  in 164 

Cylinders,  strength  of 138 

,  thick 144 

Eccentrics 174 

Eccentric  rods 189 

straps 176 

Elasticity 1 

Expansion  joints 143 

Factor  of  safety 1 

Fly-wheels 155 

,  construction  of 161 

,  design  of 155 

?  stresses  in ............................................  159 


INDEX  243 

,  PAGE 

Friction  wheels 107 

,  bevel 108 

,  grooved 109 

,  methods  of  engaging 109 

Gears,  bevel 116 

,  rims  and  arms  of 119 

,  spiral 117 

,  split 120 

,  spur Ill 

,  strength  of  teeth  of Ill 

,  worm 117 

Hooks 206 

Iron 224 

,  cast 225 

,  malleable 226 

,  wrought 226 

Journals 75 

,  heating  of 76 

Keys,  kinds  of 46 

,  proportions  of 49 

,  stresses  in 47 

Knuckle  thread 22 

Lap  joint,  double  riveted 39 

,  single  riveted 38 

,  triple  riveted 39 

Lewis  equation 115 

Load 1 

Locking  devices 24 

Lubrication  of  bearings 79 

Modulus  of  electricity 2 

Pipe  joints 140 

Pipes,  material  and  manufacture 137 

,  size  of 145 

,  strength  of 138 

Piston  rings 193 

Piston  rods. .  .  .  , 187 

Pistons .191 


244  INDEX 

PAGE 

Pulleys 101 

,  cone 104 

,  proportions  of  arms,  etc 101 

,  split 103 

,  tight  and  loose 103 

Ratchets 214 

Riveted  joints,  design  of 43 

,  strength  of 34 

,  types  of 32 

Rivets,  types  of 32 

Rope  driving,  systems  of 121 

Ropes,  materials 121 

,  power  transmission 125 

,  pulleys , 124 

Screws,  machine 24 

,  set 24 

Screw  threads,  table  of 20 

Screws,  transmission 28 

Section  moduli,  table  of 10 

,  polar,  table  of 16 

Shafts,  hollow. 58 

,  practical  rules  for  diameters  of 60 

subjected  to  bending  and  twisting 59 

subjected  to  torsion  only 57 

,  torsional  rigidity  of 61 

Shear 5 

Shrink  and  force  fits 54 

links.: 54 

Springs 216 

,  coil 218 

,flat 216 

,  leaf 217 

,  material  of 220 

,  spiral 218 

Square  thread 21 

Steel 228 

,  alloy 229 

,  Bessemer 228 

,  high  carbon 229 

,  mild 228 

,  open  hearth 228 

for  bearings 78 

1 

Stresses,  combined 15 


INDEX  245 

PAGE 

Stresses,  compound 6 

Stuffing  boxes 199 

Tension 2 

Timber 233 

Torsion 14 

U.  S.  standard  thread 19 

Valves,  lift 151 

,  types  of 148 

V  thread 21 

White  metal  alloys 232 

Whitworth  thread 19 

Wire  rope  transmission 127 

Wrist  pin 197 


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